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2 years ago

a full-time employee who works 40 hours per week earns 29.85 per hour estimate the person's annual income

Mathematics

2 answers:

balu736 [363]2 years ago

8 0

There are 52 weeks in a year. The question is asking how much is the annual salary.

Luda [366]2 years ago

7 0

Their annual income would be around 62,088

40h x $29.85 = $1,194 per week
then take $1,194 x 52 (because there are 52 weeks in a year) = $62,088 annual income

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A rectangle has two side lengths of 8 inches and two side lengths of 10 inches. What is the perimeter of the rectangle?

Ronch [10]

The perimeter is 36
8+8=16
10+10=20
20+16=36

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Ana can finish a certain job in 10 hours. Together, Ana and Nina can finish the same job in 3 hours. How many hours can Nina fin

Over [174]

Answer: 4.5 hours

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

$Ana: \dfrac{1}{10}\ \text{job per hour}\\\\Nina: \dfrac{1}{x}\ \text{job per hour}\\\\Together: \dfrac{1}{3}\ \text{job per hour}\\\\\\Ana + Nina = Together\\\dfrac{1}{10}\quad +\dfrac{1}{x}\quad =\quad \dfrac{1}{3}\\\\\\\dfrac{1}{10}(30x)+\dfrac{1}{x}(30x) =\dfrac{1}{3}(30x)\\\\\\\rightarrow\quad 3x+30=10x\\\\.\quad 3x+30=10x\\\underline{-3x\qquad \quad -3x\quad}\\.\qquad \quad 30=7x\\\\.\qquad \quad \dfrac{30}{7}=\dfrac{7x}{7}\\\\\\.\qquad \quad \dfrac{30}{7}=x$

$.\qquad \quad 4.5=x$

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2 years ago

Is 7, 20, 25 a right triangle?

love history [14]

Answer:

The answer to your question is below

Step-by-step explanation:

Process

To answer if they are right triangles or not use the Pythagorean theorem which states that the square of the hypotenuse equals the sum of the square of the legs.

a) 25² = 20² + 7²

625 = 400 + 49

625 ≠ 449 It is not a right triangle

b) 26² = 24² + 10²

676 = 576 + 100

676 = 676 it is a right triangle

c) 13² = 8² + (√10)²

169 = 64 + 10

169 ≠ 74 it is not a right triangle

d) 52² = 48² + 20²

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2 years ago

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kherson [118]

Answer:

The answer is

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Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

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