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Misha Larkins [42]
3 years ago
7

ANSWER ALL 5 PARTS.

Mathematics
1 answer:
N76 [4]3 years ago
6 0
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}}

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Which are true or false
Savatey [412]

we are given

y=19.75x

Here,

y is the number of gallons of water in the pool

x is the number of minutes the truck has been filing the pool

now, we will verify each options

option-A:

When x=1

then y=19.75

so, for 1 minute , number gallons =19.75

so, this is FALSE

option-B:

Since, we have

y=19.75x

It is similar to equation of line

y=mx+b

But b=0

It means that it passes through origin

so, this is TRUE

option-C:

y=19.75x

we can compare it with

y=mx

where m is the rate

so, we get

m=19.75

while we are given

the water in swimming pool increase about 100 gallons by every 5 minutes

so, we get rate as

m=\frac{100}{5} =20 gallons per minute

this is close to 19.75

so, they are approximately equal

so, this is TRUE

option-d:

x-value and y-value will show relationship

and ratio will always be constant or equivalent

because it has linear relationship

so, this is TRUE

option-e:

We need to check

(8,150) ..so, x=8 and y=150

we can plug this in our equation

and we get

150=19.75\times 8

150=158

Since, left side is not equal to right side

so, this can not be point

so, this is FALSE


5 0
3 years ago
A 6-pound bag of chocolate candies costs $14.04. What is the unit price?
Ber [7]

Answer:

2.34 per pound is unit price

Step-by-step explanation:

14.04 divided by 6 is 2.34

3 0
3 years ago
A,B,C or D for the peanuts
disa [49]

Answer:    I THINK it is c

<h2>Step-by-step explanation:</h2>

7 0
3 years ago
X2 - 6x + 12 and y = 2x - 4, algebraically are
olga_2 [115]

<em><u>The solution is (4, 4)</u></em>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2

<em><u>Substitute eqn 2 in eqn 1</u></em>

x^2 - 6x + 12 = 2x - 4

Make the right side of equation 0

x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0

<em><u>Solve by quadratic equation</u></em>

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4

<em><u>Substitute x = 4 in eqn 2</u></em>

y = 2(4) - 4

y = 8 - 4

y = 4

Thus solution is (4, 4)

5 0
3 years ago
The difference of m and 2
sergejj [24]
Difference means you want to find the change between two, thus u subtract between both of it.

m-2
5 0
3 years ago
Read 2 more answers
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