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olya-2409 [2.1K]
3 years ago
6

Given the functions k(x) = 5x − 8 and p(x) = x − 4, solve k[p(x)] and select the correct answer below.

Mathematics
2 answers:
Goryan [66]3 years ago
8 0
<span>Answer is B) k[p(x)] = 5x − 28</span>
Alex787 [66]3 years ago
8 0

Answer:

The answer is B

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What is the simplify of this expression -2(2b-3)
patriot [66]

-2(2b-3)

Multiply the bracket by -2

Whenever multiplying a -negative number with a -negative number= + positive number.

(-2)(2b)(-2)(-3)

-4b+6

Answer: -4b+6

6 0
4 years ago
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Find the Domain: <br> {(-5, 3), (-1,0), (3, -4), (-1, 2)}
Marrrta [24]

Answer:

(-1,2) is domain

Step-by-step explanation:

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4 years ago
The Student Council of your school orders 11 pizzas for an event.
bezimeni [28]

Answer:

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Step-by-step explanation:

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3 years ago
I need to find the perimeter and area of the regular polygon.
mezya [45]
The area of a polygon is given by the formula Area = ap/2 where a is the length of the apothem and p is the perimeter. The apothem is a line from the center of the polygon perpendicular to a side.

Depending on the formula you know, you can find the length of a side in 1 of 2 ways. 

The first way uses a triangle. Using the radius of the polygon you can create 8 congruent triangles. The center angle will be 360 / 8 = 45 and two side lengths of 20. You can find the length of the base using the law of cosines.

c^2 = 20^2 + 20^2 - 2(20)(20)(cos 45)
c^2 = 400 + 400 - 800(cos 45)
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c = sqrt(800 - 800(cos 45)
c = 15.31

The second way is to use this formula:
r = s / (2 sin(180 / n))
20 = s / (2 sin(180/8)
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(40)sin(22.5) = s
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We need to calculate the perimeter. As there are 8 sides (8)(15.31) = 122.48
Now we need to calculate the apothem using

a = S / (2 tan (180 / n)
a = 15.31 / (2 tan (180 / 8))
a = 18.48

Now solve for the area

Area = ap/2
Area = (18.48)(122.48)/2
Area = 1131.72

perimeter = 122.48
area = 1131.72
7 0
3 years ago
What’s the slope of (-3, 6), (2, 6)?
IRINA_888 [86]

Answer:

0

Step-by-step explanation:

(y2-y1)/(x2-x1)

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2(x2)- -3(x1)=5

0/5=0

4 0
3 years ago
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