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KengaRu [80]
3 years ago
9

A rectangular pool 6 meters by 4 meters is surrounded by a walkway of width x meters. At what value of x will the area of the wa

lkway equal the area of the pool?

Mathematics
2 answers:
Vesnalui [34]3 years ago
4 0

Answer:

<h2>x = 1</h2>

Step-by-step explanation:

To answer the question we need to calculate the area of the pool and the expression of the walkway area and then make them equal.

So, the area of the pool would be:

A_{pool}=b.h=(6m)(4m)=24m^{2}

Then, the area os the walkway would be the difference between the whole are combines and the area of the pool. The whole are combined, pool plus walkway would be:

A_{total}=(x+6+x)(x+4+x)=(2x+6)(2x+4)=4x^{2}+8x+12x+24=4x^{2}+20x+24

Now, the area of the walkway is:

A_{walkway}=A_{total}-A_{pool}\\A_{walkway}=4x^{2}+20x+24-24=4x^{2}+20x

Then, we make equal the expression of the area of the walkway to the area of the pool, because the questions is asking the x-value when they are equal.

4x^{2}+20x=24\\4x^{2}+20x-24=0

Now, we solve the quadratic equation. We can extract the common factor to have an easier equation:

4x^{2}+20x-24=0\\4(x^{2}+5x-6)=0\\x^{2}+5x-6=0

Then, we have to find to number which multiplication results in 6 and their difference is 5. We find that the numbers are -6 and 1.

Therefore, the solutions are

x=-6;x=1

Then the solution is one, that is, if the width of the walkway is 1m it would have the same are than the pool. Also, the solution must be just the positive number, because negative lengths don't make sense.

steposvetlana [31]3 years ago
3 0

Answer:

x=1 meter

Step-by-step explanation:

step 1

Find the area of the rectangular pool

A=LW

we have

L=6\ m\\W=4\ m

substitute

A=6(4)=24\ m^2

step 2

Find the area of rectangular pool including the area of the walkway

Let

x ----> the width of the walkway

we have

L=(6+2x)\ m\\W=(4+2x)\ m

substitute

A=(6+2x)(4+2x)

step 3

Find the area of the walkway

To find out the area of the walkway subtract the area of the pool from the area of rectangular pool including the area of the walkway

so

A=(6+2x)(4+2x)-24

step 4

Find the value of x if the area of the walkway equal the area of the pool

so

(6+2x)(4+2x)-24=24

Solve the quadratic equation by graphing

The solution is x=1 meter

see the attached figure

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Ilia_Sergeevich [38]
X/8 + 12 = 15
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4 0
3 years ago
The mean score of 15 students was 82.0. (a) Another student joins, and the mean becomes 82.5. What is the new student's score? (
slega [8]

Answer:

a) The new student's score is 90.

b) Their scores are a pair of scores lower than 100 with a mean of 90.5.

Step-by-step explanation:

The mean is the sum of all scores divided by the number of students.

The mean score of 15 students was 82.0

This means that there are 15 students and the sum of the scores is 15*82 = 1230.

(a) Another student joins, and the mean becomes 82.5. What is the new student's score?

Now there are 16 students, the sum of the scores is 1230 + x and the mean is 82.5. So

82.5 = \frac{1230 + x}{16}

1230 + x = 16*82.5

x = 1320 - 1230

x = 90

The new student's score is 90.

(b) Suppose instead that two students join, taking the average from 82 to 83. What could their scores be, if they are restricted to be integer (i.e. whole) numbers?

Now there are 2 new students, so 17 in total. The mean score of the new students is 2x. The total score of the class is 1230 + 2x and the mean is 83. So

83 = \frac{1230 + 2x}{17}

1230 + 2x = 1411

2x = 181

x = 90.5

The mean of the two students scores is 90.5.

So their scores are a pair of scores lower than 100 with a mean of 90.5.

4 0
3 years ago
Find the Value of X (In this picture)​
slava [35]

Answer:

x = 63

Step-by-step explanation:

It's Quadrilateral so:

360-90-72= x + x +22 (Cause that 2 triangle is isosceles)

2x=126

x=63

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3 years ago
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yanalaym [24]
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6 0
3 years ago
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Find the approximate perimeter of ABC plotted below.
maksim [4K]

Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}

AB = \sqrt{4 + 4}

AB = \sqrt{16}

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}

BC = \sqrt{25 + 49}

BC = \sqrt{74}

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}

AC = \sqrt{47 + 25}

AC = \sqrt{74}

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

8 0
3 years ago
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