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3 years ago

The scatterplot below represents the number of hot drinks sold on a given

Mathematics

1 answer:

USPshnik [31]3 years ago

3 0

Answer:

A) y=-1/2x+55

Step-by-step explanation:

We can clearly see a negative trend/association between the number of hot drinks sold on a given day and the low temperature of that day. Therefore, the slope of the best line of fit must be negative, eliminating D and E. Also, notice that when the low temperature is 0°F, the number of hot drinks sold is about 55, which is a good indicator of our y-intercept or constant, eliminating choice C. Because the best line of fit is not so steep, the slope will be closer to 0 if it's a negative trend. This eliminates choice B, making the correct answer A.

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HELP<br> Factor<br> 6x^2+10x-4

Nimfa-mama [501]

Answer:

46x-4 i think

6x^2= 36x

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so 46x-4 is the answer

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Why is the answer to this integral's denominator have 1+pi^2

ss7ja [257]

It comes from integrating by parts twice. Let

$I = \displaystyle \int e^n \sin(\pi n) \, dn$

Recall the IBP formula,

$\displaystyle \int u \, dv = uv - \int v \, du$

Let

$u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Then

$\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn$

Apply IBP once more, with

$u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Notice that the ∫ v du term contains the original integral, so that

$\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)$

$\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I$

$\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n$

$\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C$

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2 years ago

A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of

Ivenika [448]

The equation for the height of the rocket at time t given

$h= -16t^2+192t$

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

$h= -16t^2+192t$

$560 = -16t^2+192t$

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

$560 = -16(t^2 - 12t)$

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

$560/-16 = -16(t^2-12t)/-16$

$-35 = t^2 -12t$

Now we will move -35 to the righ side by adding 35 to both sides.

$-35+35 = t^2-12t+35$

$0 = t^2 -12t+35$

$t^2-12t+35 = 0$

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

$t^2-12t+35 =0$

$(t-5)(t-7) =0$

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

$h= -16t^2+192t$

$0= -16t^2 + 192 t$

$0 = -16(t^2-12t)$

$-16(t^2-12t) = 0$

We will move -16 to the other side by dividing it to both sides.

$-16(t^2-12t)/-16 = 0/-16$

$t^2-12t = 0$

We will take out the common factor t from the left side. By taking out t we will get,

$t(t-12) = 0$

We will use zero product property now. By using that we will get,

$t = 0$

ans also $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

6 0

3 years ago