The function g is defined by the following rule. g (x) = 4x + 4

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!
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[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]

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

Answer:

Step-by-step explanation:

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

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

Wavelength=speed/frequency

=2/15

=0.1333 m

Bulb 2 was activated 760 times throughout the process.

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.

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:

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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]

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.

Answer:

the area of the triangle is 36

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~

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)

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

Linear approximations are used to estimate functions using derivatives

The approximated value of sin(56 degrees) is 0.8429

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

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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:

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]

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

Answer:

16 minutes

Step-by-step explanation:

multiply 20 by 80% (20 times .80)

Step-by-step explanation:

the formula for the circumference of a circle is

2×pi×r

so, B is correct : 2×pi×4 in

Answer:

7x+9

Step-by-step explanation:

10x- 3x=7x

5--4=9

so you just bring them together so 7x+9

Answer:

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

Step-by-step explanation:

Answer:

15/2 is the answer to the question

Answer:

15/2

Step-by-step explanation:

Answer:

answer is 105 hope u like it

Speed is the rate of change of distance over time

It takes Rohit 2 hours to cover the same distance

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

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(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.

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]

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

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

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.

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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