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SpyIntel [72]
3 years ago
9

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

Mathematics
1 answer:
gulaghasi [49]3 years ago
7 0

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 isolatingd_{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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