Question

(1 point)

Responses

1,280+30h=0
1 comma 280 plus 30 h is equal to 0

1,280−30h=0
1 comma 280 minus 30 h is equal to 0

10−30h=0
10 minus 30 h is equal to 0

30h=10

Answer 1

[AI-generated answer — Bot GPT-4 Turbo]

It seems you are presenting a sequence of algebraic equations. Let's solve each one for \( h \).

1. \( 1,280 + 30h = 0 \)

To solve for \( h \), we need to isolate \( h \) on one side of the equation:

\( 1,280 + 30h = 0 \)

First, subtract 1,280 from both sides:

\( 30h = -1,280 \)

Now, divide both sides by 30:

\( h = \frac{-1,280}{30} = -\frac{128}{3} \approx -42.67 \)

2. \( 1,280 - 30h = 0 \)

Similarly, we solve for \( h \) by isolating it:

\( 1,280 - 30h = 0 \)

Subtract 1,280 from both sides:

\( -30h = -1,280 \)

Divide both sides by -30:

\( h = \frac{1,280}{30} = \frac{128}{3} \approx 42.67 \)

3. \( 10 - 30h = 0 \)

Again, isolate \( h \):

\( 10 - 30h = 0 \)

Subtract 10 from both sides:

\( -30h = -10 \)

Divide by -30:

\( h = \frac{10}{30} = \frac{1}{3} \approx 0.33 \)

4. \( 30h = 10 \)

For this equation, \( h \) is already isolated on one side, so just divide by 30:

\( h = \frac{10}{30} = \frac{1}{3} \approx 0.33 \)

Now you have the values of \( h \) for each equation.