Question
Question 4 A) The function h(t)=−16t2+32th(t)=−16t2+32t gives the height of a ball kicked into the air from the ground, t seconds after it has been kicked. What would be an appropriate domain for this function? (1 point) Responses t≤16t≤16t is less than or equal to 16 t>2t>2t is greater than 2 t<16t<16t<16t<16 0≤t≤20≤t≤20≤t≤20≤t≤2 Question 5 A) The area of a rectangle is 15x2+14x−815x2+14x−8. The length of the rectangle is 3x+43x+4. What is the width of the rectangle? (1 point) Responses 12x+1212x+1212 x plus 12 5x−25x−25 x minus 2 5x+12
Answers
GPT-4o mini
### Question 4
For the function \( h(t) = -16t^2 + 32t \), which describes the height of a ball after \( t \) seconds, we need to consider the physical context. The height of the ball cannot be negative, and it would hit the ground when \( h(t) = 0 \).
Setting \( h(t) = 0 \):
\[
-16t^2 + 32t = 0
\]
Factoring out \( -16t \):
\[
-16t(t - 2) = 0
\]
This gives us \( t = 0 \) or \( t = 2 \). Since the ball is kicked from the ground (height = 0) at \( t = 0 \) and falls back to the ground at \( t = 2 \), an appropriate domain for this function is \( 0 \leq t \leq 2 \).
**Correct Answer: \( 0 \leq t \leq 2 \)**
### Question 5
To find the width of the rectangle, we can use the formula for the area of a rectangle, which is:
\[
\text{Area} = \text{Length} \times \text{Width}
\]
Given:
- Area: \( 15x^2 + 14x - 8 \)
- Length: \( 3x + 4 \)
Let the width be \( w \). The equation becomes:
\[
(3x + 4) \times w = 15x^2 + 14x - 8
\]
Now, to find \( w \), we can divide the area by the length:
\[
w = \frac{15x^2 + 14x - 8}{3x + 4}
\]
We can perform polynomial long division to simplify this:
1. Divide the leading term of the numerator \( 15x^2 \) by the leading term of the denominator \( 3x \):
\[
15x^2 \div 3x = 5x
\]
2. Multiply \( 5x \) by the entire denominator \( 3x + 4 \):
\[
5x(3x + 4) = 15x^2 + 20x
\]
3. Subtract this result from the original polynomial:
\[
(15x^2 + 14x - 8) - (15x^2 + 20x) = 14x - 20x - 8 = -6x - 8
\]
4. Now, divide \( -6x \) by \( 3x \) to get \( -2 \).
5. Multiply \( -2 \) by \( 3x + 4 \):
\[
-2(3x + 4) = -6x - 8
\]
6. Subtract this from \( -6x - 8 \):
\[
(-6x - 8) - (-6x - 8) = 0
\]
Thus:
\[
w = 5x - 2
\]
The width of the rectangle is \( 5x - 2 \).
**Correct Answer: \( 5x - 2 \)**
For the function \( h(t) = -16t^2 + 32t \), which describes the height of a ball after \( t \) seconds, we need to consider the physical context. The height of the ball cannot be negative, and it would hit the ground when \( h(t) = 0 \).
Setting \( h(t) = 0 \):
\[
-16t^2 + 32t = 0
\]
Factoring out \( -16t \):
\[
-16t(t - 2) = 0
\]
This gives us \( t = 0 \) or \( t = 2 \). Since the ball is kicked from the ground (height = 0) at \( t = 0 \) and falls back to the ground at \( t = 2 \), an appropriate domain for this function is \( 0 \leq t \leq 2 \).
**Correct Answer: \( 0 \leq t \leq 2 \)**
### Question 5
To find the width of the rectangle, we can use the formula for the area of a rectangle, which is:
\[
\text{Area} = \text{Length} \times \text{Width}
\]
Given:
- Area: \( 15x^2 + 14x - 8 \)
- Length: \( 3x + 4 \)
Let the width be \( w \). The equation becomes:
\[
(3x + 4) \times w = 15x^2 + 14x - 8
\]
Now, to find \( w \), we can divide the area by the length:
\[
w = \frac{15x^2 + 14x - 8}{3x + 4}
\]
We can perform polynomial long division to simplify this:
1. Divide the leading term of the numerator \( 15x^2 \) by the leading term of the denominator \( 3x \):
\[
15x^2 \div 3x = 5x
\]
2. Multiply \( 5x \) by the entire denominator \( 3x + 4 \):
\[
5x(3x + 4) = 15x^2 + 20x
\]
3. Subtract this result from the original polynomial:
\[
(15x^2 + 14x - 8) - (15x^2 + 20x) = 14x - 20x - 8 = -6x - 8
\]
4. Now, divide \( -6x \) by \( 3x \) to get \( -2 \).
5. Multiply \( -2 \) by \( 3x + 4 \):
\[
-2(3x + 4) = -6x - 8
\]
6. Subtract this from \( -6x - 8 \):
\[
(-6x - 8) - (-6x - 8) = 0
\]
Thus:
\[
w = 5x - 2
\]
The width of the rectangle is \( 5x - 2 \).
**Correct Answer: \( 5x - 2 \)**