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GenaCL600 [577]
2 years ago
15

HELPPP ASAP PLEASEEE THANK YOU

Mathematics
1 answer:
ahrayia [7]2 years ago
5 0

Step 1: Find the slope

Slope formula = (y2 - y1) / (x2 - x1)

---You can choose any two points when finding the slope!

Point 1 = (-2,-10)

Point 2 = (-1,-3)

Slope = (-3 - - 10) / (-1 - - 2)

Slope = 7 / 1

Slope = 7

Step 2: Find the y-intercept

Now that we have the slope, we'll plug that and one point into slope-intercept form. Then, we'll solve for b.

Slope-Intercept Form: y = mx + b

---m is the slope, b is the y-intercept

Point = (-1, -3)

Slope = 7

-3 = 7(-1) + b

-3 = -7 + b

b = 4

Step 3: Put it all together

Now that we know the slope and y-intercept, all that's left to do is plug them into slope-intercept form.

Slope-Intercept Form: y = mx + b

Slope = 7

Y-Intercept = 4

Line: y = 7x + 4

Correct Answer: A

Hope this helps!

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Read 2 more answers
ANSWER ALL 5 PARTS.
N76 [4]
A function f from a set A to a set B is defined as a relation that assings to each element  x in the set A exactly one element y in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is x feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let x be the width of the tomato patch, then the function that matches this statement is:

\boxed{Wa(x)=\frac{x}{2}}

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants <span>its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

</span>\boxed{La(x)=x+2}
<span>
2. Ar</span>ea of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then t<span>he area of a rectangle is the product of the length and width. So:

</span>A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}
<span>
3. C</span><span>ombined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

</span>Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}
<span>
4. W</span>rite the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:<span>

</span>Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x

5. 
Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that <span>Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

</span>31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06
<span>
Therefore the area of the remaining space is:

</span>\boxed{Aa(7.06)=7.06ft^{2}}

6 0
3 years ago
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