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sesenic [268]
3 years ago
6

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS

Mathematics
2 answers:
OLEGan [10]3 years ago
7 0
Your answer should be the 3rd box and 6th box for the question.
Setler79 [48]3 years ago
6 0

The answer is: C, and D

C: because the question says up to 80lb.

D: because it says at least 4 small boxes

Hope this helps!

--J.M

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What is the solution of the system of equations? 4x-3y=-1 3x-9y=33
ehidna [41]

Step-by-step explanation:

4x+13y=-9y+3y=33

17x=-6y=33

5 0
3 years ago
Solve for y in the given equation 2y - 24 = 14y <br> Answer are <br><br> 8y <br> -8y<br> -2 <br> 2
ArbitrLikvidat [17]

Answer:

y = -2

Step-by-step explanation:

First, we subtract 2y on both sides:

-24 = 12y

Then, we divide 12 on both sides:

y = -2

And we're done ^^ hope this helps!

6 0
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X+y=4 and x-y=6 substitution
vlabodo [156]
Y = -1
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Y'all is so nice too help me I really appreciate it help me thanks<br> plz i need help
butalik [34]
You can multiply 2/3 by 2/2 to get 4/6. This means that x=4!!!

Hope this helps :)
8 0
3 years ago
If x = a cosθ and y = b sinθ , find second derivative
Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

\dfrac{\mathrm d^2y}{\mathrm dx^2}

Compute the first derivative. By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta

x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta

and so

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta

Now compute the second derivative. Notice that \frac{\mathrm dy}{\mathrm dx} is a function of \theta; so denote it by f(\theta). Then

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}

By the chain rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta

and so the second derivative is

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta

4 0
3 years ago
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