What is the length of ONE side in a square if the perimeter is 4x + 25

Answers

Answer 1

Answer:

Hi, you already asked this question, the answer is 6...

Step-by-step explanation:

Answer 2
The answer is 6 I believe

Related Questions

Select the correct answer.
Which statement is true about this equation?

3(-y + 7) = 3(y + 5) + 6

A.
The equation has one solution, y = 0.
B.
The equation has one solution, y = -1.
C.
The equation has no solution.
D.
The equation has infinitely many solutions.

Answers

Answer:

one answer ( y=0 )

Step-by-step explanation:

solve for y

-3y + 21 = 3y + 15 + 6

-3y +21 = 3y +21

(-6y = 0)/-6

y=0

Answer:

A. The equation has one solution, y = 0.

Explanation:

3(-y + 7) = 3(y + 5) + 6

-3y + 21 = 3y + 15 + 6

-3y -3y = 21 - 21

-6y = 0

y = 0

There are 380 light bulbs lined up in a row in a long room.
Each bulb has its own switch and is currently switched off.
Each bulb is numbered consecutively from 1 to 380. You
first flip every switch. You then flip the switch on every
second bulb(turning off 2, 4,6...). You then flip the switch
on every third bulb (3, 6, 9...). This continues until you
have gone through the process 380 times.

Answers

Bulb 2 was activated 760 times throughout the process.

How to calculate how many times light bulb #2 was activated?

To calculate how many times light bulb #2 was activated, we must identify how many times it is activated in each process.

Based on the information provided, bulb two has the following activity.

Power on oncePower off once

According to the above, bulb #2 is activated 2 times each process. So to know how many times it is activated in total, we must multiply the number of times it is activated in each process by the total number of processes (380).

380 × 2 = 760

Note: This question is incomplete because the question is missing. Here is the question:

After repeating this process 380 times, how many times was light bulb 2 activated?

Learn more about light bubs in: https://brainly.com/question/4723473

Determine the hcf and Lcf of 1820 and 3510​

Answers

Answer:

To get the Least Common Multiple (LCM) of 1820 and 3510 we need to factor each value first and then we choose all the factors which appear in any column and multiply them:

1820: 2 2 5 7 13

3510: 2 3 3 3 5 13

LCM: 2 2 3 3 3 5 7 13

The Least Common Multiple (LCM) is: 2 x 2 x 3 x 3 x 3 x 5 x 7 x 13 = 49140

Find the value of N in 26%×N=78​

Answers

Answer:

26%×N=78

Reduce the fraction 26/100 to lowest terms by extracting and cancelling out 2.

13

N=78

50

multiply both sides by 15/30 the reciprocal of 13/50

N=78× (50/13)

Express 78×(50/13) as a single fraction

N=78×50

13

multiply 78 and 50 to get 3900.

N=3900

13

Divide 3900 by 13 to get 300.

N=300

So the answer is N=300

Step-by-step explanation:

#Carry on learning

7x - 5y = -24
-9x + 5y = 18

Answers

Answer:

Nothing further can be done with this topic. Please check the expression entered or try another topic.

7x−5y=−24−9x+5y=187x-5y=-24-9x+5y=18

Answer:

Assuming this is a system of equations...

Point form > (3, 9)

Equation form > x = 3, y = 9

Step-by-step explanation:

So we need to solve for x in 7x - 5y = -24

Add 5y to both sides

7x = -24 + 5y

-9x + 5y = 18

Divide each term by 7

7x/7 = -24/7 + 4y/7

-9x + 5y = 18

x = -24/7 + 5y/7

-9x + 5y = 18

Now we need to replace all occurences of x with -24/7 + 5y/7

-9(-24/7 + 5y/7) + 5y = 18

x = -24/7 + 5y/7

So lets focus on simplifying -9(-24/7 + 5y/7) + 5y

Apply the distributive property

-9(-24/7) - 9 5y/7 + 5y = 18

Now multiply -9(-24/7)

So -1 by -9

9(24/7) - 9 5y/7 + 5y = 18

Combine 9 and 24/7

9 * 24/7 - 9 5y/7 + 5y = 18

Then Multiply 9 by 24

216/7 - 9 5y/7 + 5y = 18

Now we multiply -9 5y/7

So Combine -9 and 5y/7

216/7 + -9(5y)/7 + 5y = 18

Now Multiply 5 by -9

216/7 + -45y/7 + 5y = 18

Move the negative

216/7 - 45y/7 + 5y = 18

Now we need to multiply by 7/7 to make 5y a fraction with a common denom.

216/7 - 45y/7 + 5y * 7/7 = 18

Combine

216/7 + -45y + 5y * 7/7 = 18

Combine further

216 - 45y + 5y * 7/7 = 18

Multiply

216 - 45y + 35y

Add

216 - 10y/7 = 18

Factor 2 out of the equation

2(108) - 10y/7 = 18

Factor more

2(108) + 2(-5y)/7 = 18

Factor further

2(108 - 5y)/7 = 18

Now we want to solve for y in 2(108 - 5y)/7 = 18

Multiply both sides by 7 then simplify.

2(108 - 5y) * 7/7 = 18 * 7

2 * 108 + 2 (-5y) = 18 * 7

Multiply

216 + 2 (-5y) = 18 * 7

Multiply again

216 - 10y = 18 * 7

Reorder 216 and -10y

-10y + 216 = 18 * 7

Simplify the right side

-10y + 216 = 126

Now we need to solve for y

So lets move all terms not containing y to the right side.

-10y = 126 - 216 (Subtract 216 from both sides)

-10y = -90

Divide each term by -10

-10y/-10 = -90/-10

Simplify the left side

-90/10

And the right side

y = 9

x = -24/7 + 5y/7

Now replace y with 9

-24/7 + 5(9)/7

Simplify the right side

-24 + 5(9)/7

Multiply

-24 + 45

Add

21/7

x = 3

Therefore x = 3, y = 9 > (3, 9)

PLEASE HELP I'LL DO ANYTHING


Simplify the expression

1/5 (5x + 9) + 4/5 (1 - 9x)​

Answers

Step-by-step explanation:

Use the Distributive Property:

[tex]1/5(5x+9)+4/5(1-9x)[/tex]

[tex]x+1.8+4/5-7.2x[/tex]

Combine Like-Terms:

[tex]-6.2x+2.6[/tex]

(1 point) Use differentials (or equivalently, a linear approximation) to approximate sin(56∘)
sin(56∘) as follows: Let ()=sin() and find the equation of the tangent line to () at a "nice" point near 56∘. Then use this to approximate sin(56∘).
Approximation =

Answers

Linear approximations are used to estimate functions using derivatives

The approximated value of sin(56 degrees) is 0.8429

How to approximate sin(56)

The trigonometry expression is given as:

[tex]\sin(56^o)[/tex]

Convert 56 degrees to radians

[tex]56^o = \frac{56}{180}\pi[/tex]

To approximate, we make use of 45 degrees.

Where:

[tex]\sin(45^o) = \cos(45^o) = \frac{\sqrt 2}{2}[/tex]

Also, we have:

[tex]45^o= \frac{\pi}{4}[/tex]

And

[tex](\sin\ x)'= \cos\ x[/tex]

So, the approximation of sin(56 degrees) become:

[tex]\sin(56\°) = \sin(45\°) + (\frac{56}{180}\pi - \frac{\pi}{4}) *\cos(45\°)[/tex]

Substitute known values

[tex]\sin(56\°) = \frac{\sqrt 2}{2} + (\frac{56}{180}\pi - \frac{\pi}{4}) *\frac{\sqrt 2}{2}[/tex]

Take LCM

[tex]\sin(56\°) = \frac{\sqrt 2}{2} + \frac{56 - 45}{180}\pi *\frac{\sqrt 2}{2}[/tex]

[tex]\sin(56\°) = \frac{\sqrt 2}{2} + \frac{11}{180}\pi *\frac{\sqrt 2}{2}[/tex]

Solve the expression

[tex]\sin(56^o) = 0.8429[/tex]

Hence, the approximated value of sin(56 degrees) is 0.8429

Read more about linear approximation at:

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find the smallest possible value of n for which 99n is multiple of 24​

Answers

Answer:

99 is not multiple of 24 you will get it wrong if you think it is

Step-by-step explanation:

Answer:

The answer is 8.

It takes Jada 20 minutes to walk to school. It takes Andre 80 percentage as long to walk to school.How long does it take Andre to walk to school?

Answers

Answer:

16 minutes

Step-by-step explanation:

multiply 20 by 80% (20 times .80)

6.Name two streets that intersect.

7.Name two streets that are parallel

Answers

Answer:

6.Elm and Oak intersect

7.Birch and Maple are parallel

Step-by-step explanation:

If Elm and Oak continue, they will intersect with each other

Birch and Maple have same distance consistently between them

Rajeev started to move from point A towards point B exactly an hour after Rohit started from B in the opposite direction but at a speed twice as much as that of rohit.By the time rohit covers ⅙ of the distance between the point B and A Rajeev also covers the same distance.

Answers

Speed is the rate of change of distance over time

It takes Rohit 2 hours to cover the same distance

How to determine the time

Represent Rajeev with A, and Rohit with B

Speed is calculated as:

[tex]Speed = \frac{Distance}{Time}[/tex]

Rohit covers 1/6 of the distance between point AB

So, we have:

[tex]S_B = \frac{AB/6}{T_B}[/tex]

Make T the subject

[tex]T_B = \frac{AB/6}{S_B}[/tex]

[tex]T_B = \frac{AB}{6S_B}[/tex]

Rajeev's speed is twice that of Rohit.

So, we have:

[tex]S_A = 2 * S_B[/tex]

[tex]S_A = \frac{AB}{T_A}[/tex]

So, the time taken by Rajiv to cover 1/6 of the distance is:

[tex]T_A = \frac{AB}{12S_B}[/tex]

The difference between the time is given as 1.

So, we have:

[tex]\frac{AB}{6S_B} - \frac{AB}{12S_B} = 1[/tex]

Multiply through by 12SB

[tex]2AB - AB = 12S_B[/tex]

[tex]AB = 12S_B[/tex]

Recall that:

[tex]T_B = \frac{AB}{6S_B}[/tex]

So, we have:

[tex]T_B =\frac{12S_B}{6S_B}[/tex]

[tex]T_B = 2[/tex]

Hence, it takes Rohit 2 hours to cover the same distance

Read more about speed at:

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Find the difference between 3x+5 and 10x-4.

Answers

Answer:

7x+9

Step-by-step explanation:

10x- 3x=7x

5--4=9

so you just bring them together so 7x+9


Is anybody able to help me with this?

Answers

Answer:

Step-by-step explanation:

How can you use
ratios to determine if a
relationship is
proportional?

Answers

Proportions are corresponding assuming they address a similar relationship. One method for checking whether two proportions are corresponding is to keep in touch with them as divisions and afterward decrease them. Assuming the decreased divisions are something similar, your proportions are relative.

Answer:

Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.

Step-by-step explanation:

Consider five circles with radii of 1, 2, 4, 8, and 16 inches.

a. Complete the table.
b. Compare the areas and circumferences. What happens to the circumference of a circle when you double the radius? What happens to the area?
c. What happens when you triple the radius?

Please answer all questions or just the table, because I need help. Thanx!

Answers

Answer:

a) 2. 4pi (in) , 4pi  

  3. 8pi , 16pi

  4. 16pi, 64pi

  5. 32 pi  ,  256pi

Step-by-step explanation:

b) when radius increase , the areas and circumferences increase to

circumference = 2 pi * radius ; so if you double the radius , circumference will be double

area = pi * radius * radius ; if you double the radius , area will be 2^2 or 4 times

c) circumference will be triple and

area will be 3^2 or 9 times

which systems have infinite solution? check all that apply
A. y = 0.5x + 2.75 and 2y = x + 2.75
B. y = 0.5x + 2.75 and y - 0.5x = 2.75
C. y = 0.5x + 2.75 and 0.5x + y = 2.75
D. y = 0.5x + 2.75 and y = 0.5(x + 5.5)
E. y = 0.5x + 2.75 and y = -2(-0.25x) + 2.75​

Answers

Answer:

B

Step-by-step explanation:

Answer:

Ok got it

Step-by-step explanation:

the answers are y = 0.5x + 2.75 and y – 0.5x = 2.75

y = 0.5x + 2.75 and y = 0.5(x + 5.5)

y = 0.5x + 2.75 and y = –2(–0.25x )+ 2.75

Section 8.1 Introduction to the Laplace Transforms

Problem 1.
Find the Laplace transforms of the following functions by evaluating the integral
[tex]F(s) = {∫}^{ \infty } _{0} {e}^{ - st} f(t)dt[/tex]
[tex](a)t[/tex]
[tex](b) {te}^{ - t} [/tex]
[tex](c) {sinh} \: bt[/tex]
[tex](d) {e}^{2t} - {3e}^{t} [/tex]
[tex](e) {t}^{2} [/tex]

Answers

For the integrals in (a), (b), and (e), you'll end up integrating by parts.

(a)

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt[/tex]

Let

[tex]u = t \implies du = dt[/tex]

[tex]dv = e^{-st} \, dt \implies v = -\dfrac1s e^{-st}[/tex]

Then

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = uv\bigg|_{t=0}^{t\to\infty} - \int_0^\infty v\, du[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = \left(-\frac1s te^{-st}\right)\bigg|_0^\infty + \frac1s \int_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1s \left(\lim_{t\to\infty} te^{-st} - 0\right) + \frac1s \int_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = \frac1s \int_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1{s^2} e^{-st} \bigg|_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1{s^2} \left(\lim_{t\to\infty}e^{-st} - 1\right) = \boxed{\frac1{s^2}}[/tex]

(b)

[tex]\displaystyle \int_0^\infty t e^{-t} e^{-st} \, dt = \int_0^\infty t e^{-(s+1)t} \, dt[/tex]

Let

[tex]u = t \implies du = dt[/tex]

[tex]dv = e^{-(s+1)t} \, dt \implies v = -\dfrac1{s+1} e^{-(s+1)}t[/tex]

Then

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\dfrac1{s+1} te^{-(s+1)t} \bigg|_0^\infty + \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\dfrac1{s+1} \left(\lim_{t\to\infty}te^{-(s+1)t} - 0\right) + \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\frac1{(s+1)^2} e^{-(s+1)t} \bigg|_0^\infty[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\frac1{(s+1)^2} \left(\lim_{t\to\infty}e^{-(s+1)t} - 1\right) = \boxed{\frac1{(s+1)^2}}[/tex]

(e)

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt[/tex]

Let

[tex]u = t^2 \implies du = 2t \, dt[/tex]

[tex]dv = e^{-st} \, dt \implies v = -\dfrac1s e^{-st}[/tex]

Then

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = -\frac1s t^2 e^{-st} \bigg|_0^\infty + \frac2s \int_0^\infty t e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = -\frac1s \left(\lim_{t\to\infty} t^2 e^{-st} - 0\right) + \frac2s \int_0^\infty t e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = \frac2s \int_0^\infty t e^{-st} \, dt[/tex]

The remaining integral is the transform we found in (a), so

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = \frac2s \times \frac1{s^2} = \boxed{\frac2{s^3}}[/tex]

Computing the integrals in (c) and (d) is much more immediate.

(c)

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \int_0^\infty \frac{e^{bt}-e^{-bt}}2 \times e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \int_0^\infty \left(e^{(b-s)t} - e^{(b+s)t}\right) \, dt[/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left(\frac1{b-s} e^{(b-s)t} - \frac1{b+s} e^{(b+s)t}\right) \bigg|_0^\infty[/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left[\lim_{t\to\infty}\left(\frac1{b-s} e^{(b-s)t} - \frac1{b+s} e^{(b+s)t}\right) - \left(\frac1{b-s} - \frac1{b+s}\right)\right][/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left(\frac1{b+s} - \frac1{b-s}\right) = \boxed{\frac{s}{s^2-b^2}}[/tex]

(d)

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \int_0^\infty \left(e^{(2-s)t} - 3e^{(1-s)t}\right) \, dt[/tex]

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \left( \frac1{2-s} e^{(2-s)t} - \frac3{1-s} e^{(1-s)t} \right) \bigg|_0^\infty[/tex]

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \lim_{t\to\infty} \left( \frac1{2-s} e^{(2-s)t} - \frac3{1-s} e^{(1-s)t} \right) - \left( \frac1{2-s} - \frac3{1-s} \right)[/tex]

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \frac3{1-s} - \frac1{2-s} = \boxed{-\frac{2s-5}{s^2-3s+2}}[/tex]



Find the area and circumference of this circle. Write your answer correct to the nearest hundredth

Answers

Answer:

Step-by-step explanation:

So we have that the diameter is 30, meaning the radius is 15.

Area: [tex]A=\pi r^{2}=\pi \cdot 15^{2}=225\pi \approx 706.86[/tex]

Circumference: [tex]C=2\pi r=2\pi \cdot 15 = 30\pi \approx 94.25[/tex]

In how many ways can first, second, and third prizes be awarded in a contest with 600 contestants?

Answers

Answer:

214,921,200 ways

Step-by-step explanation:

[tex]first \: price \: to \: one \: of \: the \: 600 \: \\ participant \\ so \: 600 \: choices \\ 2nd \: 599 \: choices \\ 3rd \: 598 \: choices \\ simplify \: 600 \times 599 \times 598 = \\ = 214,921,200[/tex]

Answer:

214,921,200 ways

Step-by-step explanation:

First = 600

Second = 599

Third = 598

Hence,

600 × 599 × 598

359400 × 598

214,921,200

Hence, in 214,921,200 different ways.

~Lenvy~

Calculus AB Homework, does anyone know how to do this...

Answers

(a) f(x) is continuous at x = 1 if the limits of f(x) from either side of x = 1 both exist and are equal:

[tex]\displaystyle \lim_{x\to1^-}f(x) = \lim_{x\to1} (2x-x^2) = 1[/tex]

[tex]\displaystyle \lim_{x\to1^+}f(x) = \lim_{x\to1} (x^2+kx+p) = 1 + k + p[/tex]

So we must have 1 + k + p = 1, or k + p = 0.

f(x) is differentiable at x = 1 if the derivative at x = 1 exists; in order for the derivative to exist, the following one-sided limits must also exist and be equal:

[tex]\displaystyle \lim_{x\to1^-}f'(x) = \lim_{x\to1^+}f'(x)[/tex]

Note that the derivative of each piece computed here only exists on the given open-ended domain - we don't know for sure that the derivative *does* exist at x = 1 just yet:

[tex]f(x) = \begin{cases}2x-x^2 & \text{for }x\le1 \\ x^2+kx+p & \text{for }x>1\end{cases} \implies f'(x) = \begin{cases}2 - 2x & \text{for }x < 1 \\ ? & \text{for }x = 1 \\ 2x + k & \text{for }x > 1 \end{cases}[/tex]

Compute the one-sided limits of f '(x) :

[tex]\displaystyle \lim_{x\to1^-}f'(x) = \lim_{x\to1} (2 - 2x) = 0[/tex]

[tex]\displaystyle \lim_{x\to1^+}f'(x) = \lim_{x\to1} (2x+k) = 2 + k[/tex]

So if f '(1) exists, we must have 2 + k = 0, or k = -2, which in turn means p = 2, and these values tell us that we have f '(1) = 0.

(b) Find the critical points of f(x), where its derivative vanishes. We know that f '(1) = 0. To assess whether this is a turning point of f(x), we check the sign of f '(x) to the left and right of x = 1.

• When e.g. x = 0, we have f '(0) = 2 - 2•0 = 2 > 0

• When e.g. x = 2, we have f '(2) = 2•2 - 2 = 2 > 0

The sign of f '(x) doesn't change as we pass over x = 1, so this critical point is not a turning point. However, since f '(x) is positive to the left and right of x = 1, this means that f(x) is increasing on (-∞, 1) and (1, ∞).

(c) The graph of f(x) has possible inflection points wherever f ''(x) = 0 or is non-existent. Differentiating f '(x), we get

[tex]f'(x) = \begin{cases}2-2x & \text{for }x<1 \\ 0 & \text{for }x=1 \\ 2x+k & \text{for }x>1\end{cases} \implies f''(x) = \begin{cases}- 2 & \text{for }x < 1 \\ ? & \text{for }x = 1 \\ 2 & \text{for }x > 1 \end{cases}[/tex]

Clearly f ''(x) ≠ 0 if x < 1 or if x > 1.

It is also impossible to choose a value of f ''(1) that makes f ''(x) continuous, or equivalently that makes f(x) twice-differentiable. In short, f ''(1) does not exist, so we have a single potential inflection point at x = 1.

From the above, we know that f ''(x) < 0 for x < 1, and f ''(x) > 0 for x > 1. This indicates a change in the concavity of f(x), which means x = 1 is the only inflection point.

4) In addition to using graphs, you should also be able to use a table to determine values in an equation. Fill in the following table based on your equation in question three defined the cost for these amount of loaves of bread that you buy.

(Equation is question 3 is y=1.60x+0)

5) where would you break even? In other words, how many loaves of bread would you need to buy where it would be the same cost to make those loaves of bread yourself? Look for the intersection of the two equations when you graph them on desmos.
How much would it cost to buy that many loaves of bread?

6) Give three advantages and three disadvantages to buying a bread maker. Would you choose to buy the bread maker or not? How did you make your decision?

Answers

It would take 125 loaves at a cost of $200 for the breadmaker and store bought bread to cost the same.

Linear equation

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the rate of change and b is the y intercept.

Let x represent the rate of cost of one loaf and y represent the total cost, hence:

y = 0.8x + 100

The rate of cost of one loaf is $0.8 and the start up cost is $100.

For the second bread it is given by:

y = 1.6x

For both cost to be the same:

1.6x = 0.8x + 100

x = 125

a = 1.6(4) = $6.4

b = 1.6(8) = $12.8

c = 1.6(12) = $19.2

It would take 125 loaves at a cost of $200 for the breadmaker and store bought bread to cost the same.

Find out more on Linear equation at: https://brainly.com/question/13763238

What is the value of triangle?

Answers

Answer:

the area of the triangle is 36

The weight of a basketball is normally distributed with a mean of 17oz and a standard deviation of 2oz.
Suppose 500 different basketballs are in a warehouse. About how many basketballs weigh more than 19oz?

O 20
O 40
O 80
O 100

Answers

80 is the correct answer ! (:

Help with math?Please? ANYONE

Answers

Answer:

(-1, 1)

Step-by-step explanation:

Hi there!

We want to solve the system of equations given as:
2x-3y=-5

3x+y=-2

Let's solve this equation by substitution, where we will set one variable equal to an expression containing the other variable, substitute the expression as the variable that it equals, solve for the other variable (the variable that the expression contains), and then use the value of the solved variable to find the value of the first variable
In the second equation, we have y by itself; therefore, if we subtract 3x from both sides, then we will get an expression that y is equal to.

So subtract 3x from both sides

y=-3x-2

Now substitute -3x-2 as y in the first equation.
It will look something like this:

2x - 3(-3x-2)=-5

Now do the distributive property.

2x+9x+6=-5

Combine like terms

11x+6=-5

Subtract 6 from both sides

11x=-11

Divide both sides by 11

x=-1

Now substitute -1 as x in the equation y=-3x-2 to solve for y:

y=-3(-1)-2

multiply

y=3-2

Subtract

y=1

The answer is x=-1, y=1; this can also be written as an ordered pair, which would be (-1, 1)
Hope this helps!
If you would like to see another problem for additional practice, take a look here: https://brainly.com/question/19212538

[tex]\begin{cases} 2x-3y=-5\\ 3x+y=-2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 2x-3y=-5\implies 2x=3y-5\implies x=\cfrac{3y-5}{2} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 2nd equation}}{3\left( \cfrac{3y-5}{2} \right)+y=-2}\implies \cfrac{3(3y-5)}{2}+y=-2 \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( \cfrac{3(3y-5)}{2}+y \right)}=2(-2)\implies 3(3y-5)+2y=-4 \\\\\\ 9y-15+2y=-4\implies 11y-15=-4\implies 11y=11[/tex]

[tex]y=\cfrac{11}{11}\implies \blacktriangleright y=1 \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{3y-5}{2}}\implies x=\cfrac{3(1)-5}{2}\implies x=\cfrac{-2}{2}\implies \blacktriangleright x=-1 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (-1~~,~~1)~\hfill[/tex]

The radius of a circle is 4 inches. How would you calculate the circumference?
A. π · 4² in.²
B. 2 · π· 4 in.
C. π · 8² in.²
D. 2· π · 8 in.

Answers

Step-by-step explanation:

the formula for the circumference of a circle is

2×pi×r

so, B is correct : 2×pi×4 in

b) 2 · π· 4 in.

Circumference of a circle: 2 • π • r

B. 2 · π· 4 in.

what is the wavelength of a wave that has a frequency of 15 Hz and a speed of 2 m/s?​

Answers

Wavelength=speed/frequency

=2/15

=0.1333 m

A recipe of beef stew serves 2 people and calls for 0.75 pounds of carrots. How many pounds of carrots would you need to serve 10 people in the restaurant? Explain how you found your answer.

Answers

Answer:

3.75 pounds of carrots

Step-by-step explanation:

Multiply 0.75 by 5 and get 3.75

Answer:

3.75 lbs of carrots

Step-by-step explanation:

10/2=5

.75*5=3.75

You basically just need to multiply your serving by 5

express 7 1/2% to fraction

Answers

Answer:

15/2 is the answer to the question

Answer:

15/2

Step-by-step explanation:

(1−x21​)5)=?
can someone help me ​

Answers

Answer:

answer is 105 hope u like it

9. Suppose you are comparing frequency data for two different
groups, 25 managers and 150 blue collar workers. Why would a
relative frequency distribution be better than a frequency
distribution?
ents

Answers

Answer:

A relative frequency distribution is better for comparison between groups whose numbers are different, since ratios are readily comparable.

Step-by-step explanation:

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