Question

question 1
lesson 5-15 practice (im g7)
a. \\(\frac{2}{5}t = 6\\)
t =
b. \\(-4.5 = a - 8\\)
a =
c. \\(\frac{1}{2} + p = -3\\)
p =
d. \\(12 = x \cdot 3\\)
x =
e. \\(-12 = -3y\\)

Explanation:

Part a

Step1: Solve for \( t \) by multiplying both sides by \( \frac{5}{2} \)

To isolate \( t \) in the equation \( \frac{2}{5}t = 6 \), we multiply both sides by the reciprocal of \( \frac{2}{5} \), which is \( \frac{5}{2} \). So we have \( t = 6\times\frac{5}{2} \).

Step2: Calculate the right - hand side

\( 6\times\frac{5}{2}=\frac{6\times5}{2}=\frac{30}{2} = 15 \)

Step1: Solve for \( a \) by adding 8 to both sides

To isolate \( a \) in the equation \( - 4.5=a - 8 \), we add 8 to both sides of the equation. So \( a=-4.5 + 8 \).

Step2: Calculate the right - hand side

\( -4.5+8 = 3.5 \)

Step1: Solve for \( p \) by subtracting \( \frac{1}{2} \) from both sides

To isolate \( p \) in the equation \( \frac{1}{2}+p=-3 \), we subtract \( \frac{1}{2} \) from both sides. So \( p=-3-\frac{1}{2} \).

Step2: Calculate the right - hand side

We can write \( - 3\) as \( -\frac{6}{2} \), then \( p=-\frac{6}{2}-\frac{1}{2}=-\frac{6 + 1}{2}=-\frac{7}{2}=-3.5 \)

Answer:

\( t = 15 \)

Part b