the table below shows the values of a function.
| x | f(x) |
| 4 | 7 |
| 6 | 11 |
| 8 | 15 |
| 10 | 19 |
which best describes the function, based on the average rates of change?
a the function is linear because the average rates of change are all the same, 2.
b the function is not linear because each average rate of change is 2 times the previous rate of change.
c the function is linear because the average rates of change are all the same, 4.
d the function is not linear because each average rate of change is 4 times the previous rate of change.
heather needs to solve the following system of equations:
equation 1: 3x - 7y = -23
equation 2: 15x + 2y = -11
she plans to eliminate the x-terms and solve for y first. which would be the best strategy for heather to use?
a multiply equation 1 by 5 and then subtract equation 2 from the result.
b multiply equation 1 by -3, and then add the resulting equations.
c multiply equation 2 by -3 and then add the result to equation 1.
d multiply equation 1 by 0 and solve equation 2 for x.

Question

Accepted Answer

### First Question (Function Description)
# Explanation:
## Step1: Recall average rate of change formula
The average rate of change between two points $(x_1, f(x_1))$ and $(x_2, f(x_2))$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$.
## Step2: Calculate rate between $x = 4$ and $x = 6$
For $x_1 = 4, f(x_1)=7$ and $x_2 = 6, f(x_2)=11$, the rate is $\frac{11 - 7}{6 - 4}=\frac{4}{2}=2$.
## Step3: Calculate rate between $x = 6$ and $x = 8$
For $x_1 = 6, f(x_1)=11$ and $x_2 = 8, f(x_2)=15$, the rate is $\frac{15 - 11}{8 - 6}=\frac{4}{2}=2$.
## Step4: Calculate rate between $x = 8$ and $x = 10$
For $x_1 = 8, f(x_1)=15$ and $x_2 = 10, f(x_2)=19$, the rate is $\frac{19 - 15}{10 - 8}=\frac{4}{2}=2$.
Since all average rates of change are 2 (constant), the function is linear.
# Answer:
A. The function is linear because the average rates of change are all the same, 2.

### Second Question (System of Equations)
# Explanation:
## Step1: Recall elimination method
To eliminate $x$-terms, we need the coefficients of $x$ in both equations to be equal in magnitude (so they can cancel when added or subtracted). The coefficient of $x$ in Equation 1 is 3, and in Equation 2 is 15.
## Step2: Analyze each option
- Option A: Multiply Equation 1 by 5: $15x-35y=-115$. Subtracting Equation 2 ($15x + 2y=-11$) would be $(15x-35y)-(15x + 2y)=-115-(-11)$, which is not eliminating $x$ (subtraction may not cancel $x$ properly here).
- Option B: Multiply Equation 1 by - 3: $-9x + 21y = 69$. Adding to Equation 2 ($15x+2y=-11$): $(-9x + 21y)+(15x + 2y)=69+(-11)$, $6x+23y = 58$ (x - terms not eliminated).
- Option C: Multiply Equation 2 by - 3: $-45x-6y = 33$. Add to Equation 1 ($3x-7y=-23$): $(3x-7y)+(-45x-6y)=-23 + 33$, $-42x-13y = 10$ (incorrect, wait no - wait, Equation 1: $3x-7y=-23$, Equation 2 multiplied by - 3: $-45x-6y = 33$. Wait, no, let's recalculate: Wait, the coefficient of $x$ in Equation 1 is 3, Equation 2 is 15. If we multiply Equation 2 by - 3, we get $-45x-6y = 33$? No, Equation 2 is $15x + 2y=-11$, multiplying by - 3: $-45x-6y = 33$. Then adding to Equation 1 ($3x-7y=-23$): $3x-7y+(-45x - 6y)=-23 + 33$ → $-42x-13y = 10$. Wait, that's not right. Wait, no, the correct way: The coefficient of $x$ in Equation 1 is 3, Equation 2 is 15. To eliminate $x$, we can make the coefficient of $x$ in Equation 1 equal to - 15 (so when added to Equation 2's 15x, they cancel). Wait, no, Equation 1: $3x-7y=-23$, Equation 2: $15x + 2y=-11$. If we multiply Equation 1 by - 5: $-15x + 35y = 115$, then add to Equation 2: $(-15x + 35y)+(15x + 2y)=115+(-11)$ → $37y = 104$, which eliminates $x$. But among the given options, Option C: Multiply Equation 2 by - 3: $-45x-6y = 33$? No, wait, no, I made a mistake. Wait, Equation 2: $15x+2y=-11$, multiply by - 3: $-45x-6y = 33$. Equation 1: $3x-7y=-23$. Wait, no, the correct approach is: The coefficient of $x$ in Equation 1 is 3, Equation 2 is 15. So if we multiply Equation 1 by 5, we get $15x-35y=-115$, then subtract Equation 2? No, better: To eliminate $x$, we can multiply Equation 1 by - 5: $-15x + 35y = 115$, then add to Equation 2 ($15x+2y=-11$): $(-15x + 35y)+(15x + 2y)=115-11$ → $37y = 104$, which eliminates $x$. But among the options, Option C: Multiply Equation 2 by - 3: $-45x-6y = 33$? No, that's not. Wait, no, let's re - evaluate the options again.

Wait, the correct way to eliminate $x$ is to have the $x$ coefficients be negatives of each other. The coefficient of $x$ in Equation 1 is 3, Equation 2 is 15. So if we multiply Equation 2 by - 3, we get $-45x-6y = 33$? No, Equation 2 is $15x + 2y=-11$, multiplying by - 3: $-45x-6y = 33$. Then Equation 1 is $3x-7y=-23$. Adding them: $3x-7y+(-45x-6y)=-23 + 33$ → $-42x-13y = 10$ (x not eliminated). Wait, I think I made a mistake in the options. Wait, Option C: "Multiply equation 2 by - 3 and then add the result to equation 1." Let's do that:

Equation 1: $3x-7y=-23$

Equation 2 multiplied by - 3: $-3\times(15x + 2y)=-3\times(-11)$ → $-45x-6y = 33$

Now add Equation 1 and the new Equation 2:

$(3x-7y)+(-45x-6y)=-23 + 33$

$3x-45x-7y-6y = 10$

$-42x-13y = 10$ (x - terms not eliminated). Wait, this is wrong. Wait, maybe I misread the options. Wait, the correct strategy is to make the coefficient of $x$ in Equation 1 equal to - 15 (so that when added to Equation 2's 15x, they cancel). So multiply Equation 1 by - 5: $-15x + 35y = 115$, then add to Equation 2 ($15x+2y=-11$): $(-15x + 35y)+(15x + 2y)=115-11$ → $37y = 104$, which eliminates $x$. But among the given options, let's re - check:

Wait, Option C: "Multiply equation 2 by - 3 and then add the result to equation 1." Wait, Equation 2: $15x+2y=-11$, multiply by - 3: $-45x-6y = 33$. Equation 1: $3x-7y=-23$. Adding them: $3x-45x-7y-6y=-23 + 33$ → $-42x-13y = 10$. No, that's not eliminating $x$. Wait, maybe the correct option is C? Wait, no, I think I made a mistake. Wait, the coefficient of $x$ in Equation 1 is 3, Equation 2 is 15. So 15 is 5 times 3. So to eliminate $x$, we can multiply Equation 1 by 5: $15x-35y=-115$, then subtract Equation 2: $(15x-35y)-(15x + 2y)=-115-(-11)$ → $-37y=-104$, which eliminates $x$. But Option A says "Multiply equation 1 by 5 and then subtract equation 2 from the result." Let's check Option A:

Equation 1 * 5: $15x-35y=-115$

Subtract Equation 2 ($15x + 2y=-11$): $(15x-35y)-(15x + 2y)=-115-(-11)$

$15x-35y-15x-2y=-104$

$-37y=-104$, which eliminates $x$. But earlier I thought it was wrong. Wait, maybe the options are different. Wait, the question is "She plans to eliminate the x - terms and solve for y first. Which would be the best strategy for Heather to use?"

Let's analyze each option:

- Option A: Multiply Equation 1 by 5: $15x-35y=-115$. Subtract Equation 2 ($15x + 2y=-11$): $(15x-35y)-(15x + 2y)=-115-(-11)$ → $-37y=-104$. This eliminates $x$ (since 15x - 15x = 0). So this works.

- Option B: Multiply Equation 1 by - 3: $-9x + 21y = 69$. Add to Equation 2 ($15x+2y=-11$): $6x+23y = 58$. $x$ not eliminated.

- Option C: Multiply Equation 2 by - 3: $-45x-6y = 33$. Add to Equation 1 ($3x-7y=-23$): $-42x-13y = 10$. $x$ not eliminated.

- Option D: Multiply Equation 1 by 0: $0x+0y = 0$, then solve Equation 2 for $x$. Not eliminating $x$ first.

Wait, so Option A is correct? But earlier I thought Option C, but I was wrong. Wait, let's recalculate Option A:

Equation 1: $3x-7y=-23$

Multiply by 5: $15x-35y=-115$ (Equation 3)

Equation 2: $15x + 2y=-11$

Subtract Equation 2 from Equation 3: $(15x-35y)-(15x + 2y)=-115-(-11)$

$15x-35y-15x - 2y=-115 + 11$

$-37y=-104$

Yes, this eliminates $x$ (15x - 15x = 0), so we can solve for $y$. So Option A is correct? But the original option A says "Multiply equation 1 by 5 and then subtract equation 2 from the result." Which is what we did. So maybe I made a mistake earlier.

Wait, the problem says "eliminate the x - terms and solve for y first". So Option A: Multiply Equation 1 by 5 (to get 15x, same as Equation 2's x - coefficient), then subtract Equation 2 (so 15x - 15x = 0), which eliminates x, allowing us to solve for y. So Option A is correct.

But wait, let's check the options again. The user's image:

Option A: "Multiply equation 1 by 5 and then subtract equation 2 from the result."

Yes, that's what we did, and it eliminates x. So the correct answer for the second question is A? Wait, no, maybe I messed up. Wait, Equation 3: $15x-35y=-115$, Equation 2: $15x + 2y=-11$. Subtracting Equation 2 from Equation 3: Equation 3 - Equation 2: $(15x-35y)-(15x + 2y)=-115-(-11)$ → $-37y=-104$, which is correct. So Option A is correct.

But I think I made a mistake earlier. So the correct answer for the second question is A.

# Answer:
A. Multiply equation 1 by 5 and then subtract equation 2 from the result.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

First Question (Function Description)

Step1: Recall average rate of change formula

Step2: Calculate rate between \(x = 4\) and \(x = 6\)

Step3: Calculate rate between \(x = 6\) and \(x = 8\)

Step4: Calculate rate between \(x = 8\) and \(x = 10\)

Step1: Recall elimination method

Step2: Analyze each option

Answer:

Second Question (System of Equations)