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[tex]d_{1}[/tex], we need to get rid of the 1/2 first. Let's multiply each side by 2:
[tex]2*m = 2*\frac{1}{2}(d_{1}+ d_{2}) \\2m = d_{1}+ d_{2}[/tex]
After this, we just subtract [tex]d_{2}[/tex] from each side to get [tex]d_{1}=2m- d_{2}[/tex]. Dark Blue is correct! Let's now plug in those numbers below:
[tex]d_{1}=2(10)- 13 = 20-13=7[/tex]
G: Let's isolate the vw^2 on one side by subtracting y from each side:
[tex]vw^{2} +y-y=x-y\\vw^{2}=x-y[/tex]
Let's now divide each side by v, then put each side under a square root to get our final answer:
[tex]\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}}[/tex]
Orange is correct! Again, let's solve the problem underneath:
w=[tex]w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3[/tex]
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.
[tex]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[/tex]
Purple is correct! Let's solve the problem:
[tex]c = 5(\frac{16}{5})+7 = 16+7 = 23[/tex]
