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2 years ago

20 POINTS!!

Mathematics

1 answer:

svet-max [94.6K]2 years ago

6 0

9514 1404 393

Answer:

P = 3x +32

Step-by-step explanation:

If we assume the short vertical edge at top center is 10 yards, as it was in two previous problems with this geometry, then we can find the area of the present parking lot as ...

(x-2)(3x) + x(3x -10) +2x(40) = 6x^2 +64x

where the left vertical rectangle is 3x high and x-2 wide; the center area between vertical rectangles is (2x-2)-(x-2) = x wide and (3x-10) high; and the right vertical rectangle is (x-2 +3x)-(2x-2) = 2x wide and 40 high.

As we said, the right vertical rectangle is 2x wide, so the area that has P as one of its dimensions is ...

A = LW = P(2x)

We want this to be the same as the existing area, so ...

2Px = 6x^2 +64x

P = 3x +32 . . . . . divide by 2x

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2 years ago

Kaitlyn's older brother borrowed money for school he took out a loan that charges 6% simple interest he will end up paying 720 a

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Step-by-step explanation:

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3 years ago

Determine whether each equation is True or False. In case you find a "False" equation, explain why is False.

elixir [45]

Answer:

(1) TRUE.

(2) FALSE.

(3) FALSE.

(4) TRUE.

(5) FALSE.

Step-by-step explanation:

(1) $\sqrt{32} = 2^{\frac{5}{2} }$

$2^{\frac{5}{2} } = (\sqrt{2} )^5 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 4\sqrt{2}\\\\\sqrt{32} = \sqrt{16 \ \times \ 2}\ = \ \sqrt{16} \ \times \ \sqrt{2} \ = \ 4\sqrt{2}$

Thus, the equation is TRUE.

(2) $16^{\frac{3}{8} } = 8^2$

$16^{\frac{3}{8} } =(2^4)^{\frac{3}{8} } = 2^\frac{3}{2} }= (\sqrt{2} )^3 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 2\sqrt{2} \\\\8^2 = 64$

Thus, the equation is FALSE.

(3) $4^{\frac{1}{2} } = \sqrt[4]{64}$

$4^{\frac{1}{2} }= \sqrt{4} = 2\\\\\sqrt[4]{64} = (64)^{\frac{1}{4} } = (2^6)^{\frac{1}{4} }= 2^{\frac{6}{4} } = 2^{\frac{3}{2} }=(\sqrt{2} )^3 = (\sqrt{2} \times \sqrt{2} \times \sqrt{2} ) = 2\sqrt{2}$

Thus, the equation is FALSE.

(4) $2^8 = (\sqrt[3]{16} )^6$

$2^8 = 256\\\\ (\sqrt[3]{16} )^6 = (16)^{\frac{6}{3} } = (2^4)^{\frac{6}{3} } = (2)^{\frac{24}{3} } = 2^8 = 256$

Thus, the equation is TRUE.

(5) $(\sqrt{64} )^{\frac{1}{3} } = 8^{\frac{1}{6} }\\\\$

$8^{\frac{1}{6} } = (2^3)^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} } = \sqrt{2} \\\\(\sqrt{64} )^{\frac{1}{3} } = (2^6)^{\frac{1}{3} } = 2^{\frac{6}{3} } = 2^2 = 4$

Thus, the equation is FALSE.

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2 years ago

Is -8 rational or irrational

ANEK [815]

Answer:

rational

Step-by-step explanation:

All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. The number 22/7 is a irrational number

Kono Dio Da!!!

6 0

3 years ago

Read 2 more answers