(1 point)

Question

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

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.