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2 years ago

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HELP! HELP! HELP! The 12 girls at Suzie’s volleyball camp have decided to wear their camp ID’s on loops of string around their n

ecks. If each girl needs 1 2 yard of string, how much string do they need in all? Suzie did this work to solve the problem: 12 1 ÷ 1 2 = 12 1 × 2 1 = 24 1 = 24 yards of string Did Suzie solve the problem correctly? No. She should have multiplied 12 and One-half to solve. No. She did not multiply the numerators correctly. No. She did not simplify 24 Over 1 correctly. Yes. Her work is correct.

1 answer:

s344n2d4d5 [400]2 years ago

6 0

Answer: No. She should have multiplied 12 and 1/2 to alive.

Step-by-step explanation:

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Hurry please I need it tonight thank u so much

Lina20 [59]

It’s the 3rd option!

-5 (the bubble shaded black), the arrow going left towards -9.

Hope this helps!

4 0

2 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.

If that's the case, then

$z_s=z_xx_s+z_yy_s$

$z_t=z_xx_t+z_yy_t$

as the hint suggests. We have

$z=\sin x\cos y\implies\begin{cases}z_x=\cos x\cos y=\cos(s+t)\cos(s^2t)\\z_y=-\sin x\sin y=-\sin(s+t)\sin(s^2t)\end{cases}$

$x=s+t\implies x_s=x_t=1$

$y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}$

Putting everything together, we get

$z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)$

$z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)$

8 0

2 years ago

Two sides of a parallelogram are 93 feet and 80 feet. The measure of the angle between these sides is 18°. Find the area of the

azamat

<h3>Answer: 2299 square feet</h3>

Work Shown:

area = a*b*sin(C)

area = 93*80*sin(18)

area = 2299.086438

area = 2299 square feet

8 0

1 year ago

Which of the expressions below are equivalent? 8x - 4 4x 4(2x - 1)

alexandr402 [8]

8x-4 and 4(2x-1)

4(2x-1) the four needs to be distributed in the parentheses.

4*2x=8x
4*(-1)=(-4)

8x-4=8x-4

4 0

2 years ago

Help me please! And quick! It's 7th grade level math...

Mademuasel [1]

Answer:

Number 5 is 10 pencils left in total(5 red, 5 blue) with 50 percent probability of getting red or blue. Number 6 is 9 left total (4 red, 5 blue) with a 44 percent probability of choosing a red one, 56 percent probability of choosing a blue one

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers