A candy bar is created with 28 cubic centimeters of chocolate. If the length is 8 cm and the width is 6 cm, what is the height of the candy bar? Round to the nearest hundredth.

Answer:

$35.00

Step-by-step explanation:

I just did it rn

Answer:

16 minutes

Step-by-step explanation:

multiply 20 by 80% (20 times .80)

Answer:

its c just so u know

Step-by-step explanation:

Answer:

the surface area of the rectangular prism is 142cm³

Step-by-step explanation:

35+15+21+21+35+15

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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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]

=====================================================

Explanation:

You can use your calculator to make short work of this problem. Simply plot the parabola and use the "minimum" feature on the calculator to spot the lowest point (specifically the coordinates of that point).

However, I have a feeling your teacher wants you to use a bit of math here. So I'll focus on another approach instead.

---------

The parabolic equation is of the form y = ax^2+bx+c

In this case, we have

Use the values of 'a' and b to find the x coordinate of the vertex h

h = -b/(2a)

h = -(-192)/(2*0.3)

h = 320

The x coordinate of the vertex is 320.

Plug this into the original equation to find the y coordinate of the vertex.

y = 0.3x^2 - 192x + 40357

y = 0.3(320)^2 - 192(320) + 40357

y = 9637 which is the minimum unit cost, in dollars.

The vertex is (h,k) = (320, 9637). It is the lowest point on this parabola.

Interpretation: If the company made 320 cars, then the unit cost (aka cost per car) is the smallest at $9637 per car.

Answer:

answer is 105 hope u like it

Answer:

v < -5

Step-by-step explanation:

Add 3 to both sides. This gives -8+(-3)>v-3+3, or -5>v. Rewriting the inequality gives v < -5.

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:

35x + 14

Step-by-step explanation:

multiply in the brackets

5x × 7 = 35x

(+)2 × 7 = (+)14

(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.

Answer:

7x+9

Step-by-step explanation:

10x- 3x=7x

5--4=9

so you just bring them together so 7x+9

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]

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.

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

Answer:

4

Step-by-step explanation:

You can set up the equation as 5x-3=2x+9. Subtract 2x from both sides, then add three to both sides which gives you 3x=12. Divide both sides by 3, and you get x=4.

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

Answer:

Step-by-step explanation:

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:

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

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

regular implies that all sides are equal
8 sides
total of 96
96/8
12

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:

The length of the other diagonal is 42 cm.

Step-by-step explanation:

Area of Rhombus = 966 cm^2 (squared)

one length of diagonal(d2) =46cm

The are of rhombus= d1 x d2/2  

*substitution from equation above* = 966= 46 x d2/2 *plug answer in* :  966=23 x d2  *divide next*

d2=966/23

d2=42cm

I don't know if that makes sense hopefully it helps.

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

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:

the area of the triangle is 36

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]

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:

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