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2 years ago

What is the constant of variation for the quadratic variation? x=2,3,4,5,6 y= 32,72,128,200,288

1 answer:

7 0

Hello,
y=a*x² ==>
32=a*2²==>a=32/4=8
72=a*3²==>a=72/9=8
128=a*4²==>a=128/16=8
200=a*5²==>a=200/25=8
288=a*6²==>a=288/36=8

Answer A: 8

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Whoever checks my work and tell what I did wrong and the right answer will

gulaghasi [49]

Answer:

See below

Step-by-step explanation:

I believe that you only had to do letters F, H, and J. In that case, let's go over each one!

F: For isolating $d_{1}$ , we need to get rid of the 1/2 first. Let's multiply each side by 2:

$2*m = 2*\frac{1}{2}(d_{1}+ d_{2}) \\2m = d_{1}+ d_{2}$

After this, we just subtract $d_{2}$ from each side to get $d_{1}=2m- d_{2}$ . Dark Blue is correct! Let's now plug in those numbers below:

$d_{1}=2(10)- 13 = 20-13=7$

G: Let's isolate the vw^2 on one side by subtracting y from each side:

$vw^{2} +y-y=x-y\\vw^{2}=x-y$

Let's now divide each side by v, then put each side under a square root to get our final answer:

$\frac{vw^2}{v} = \frac{x-y}{v}\\ w^2 = \frac{x-y}{v}\\\sqrt{w^2} = \sqrt{\frac{x-y}{v}} \\w=\sqrt{\frac{x-y}{v}}$

Orange is correct! Again, let's solve the problem underneath:

w= $w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3$

H: This one has some stuff that we haven't worked with quite yet (like terms), but our approach is the same: isolate c on one side of the equation.

$2a-2a+3c=17a-2a+21\\3c = 15a+21\\\frac{3c}{3} = \frac{15a+21}{3}\\c = \frac{15a}{3}+ \frac{21}{3} \\c = 5a+7$

Purple is correct! Let's solve the problem:

$c = 5(\frac{16}{5})+7 = 16+7 = 23$

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Answer:

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Step-by-step explanation:

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2 years ago

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anzhelika [568]

First let as solve all unit pirce
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