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3 years ago

(5-x) (b+2) using distributive properties simplify the following expression

Mathematics

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

-bx+5b-2x+10

Step-by-step explanation:

https://www.mathpapa.com/algebra-calculator.html?q=%5Cleft((5-x)(b%2B2%5Cright)

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Round 3:38 to the nearest ten

aev [14]

Answer:3

Step-by-step explanation:

Round 3.38 to the nearest ten

Rounding up to the nearest 10, we look at the digits on the left hand side of the decimal point. If the number on the right of the decimal point is less than 5, we ignore it.if it is greater than 5, we add one to the last digit on the right hand side.

In this case, the first digit on the left hand side is less than 5, so we ignore it. Rounding up to the nearest ten,the 0.38 is ignored.

Therefore, it becomes 3.0 = 3

To the nearest ten

7 0

3 years ago

if a student has homework 72 days out of 100 what fraction and what percentage of days did the student not have homework?

neonofarm [45]

Answer:

The Student has 72% out of 100%

Step-by-step explanation:

Its the same thing but with fractions

4 0

3 years ago

a young adults burns 11 calories per minute while joggers which function shows how many calories that person will burn in x minu

Zinaida [17]

Having sex is way better at burning calories, but 11x

6 0

2 years ago

How to do this and whats the answer

Vinvika [58]

Answer:

(b) 36 degrees.

Step-by-step explanation:

K is the center of the circle so triangle KMN is isosceles and m < KMN = m KNM = 26 degrees ( base angles are equal)

So m < LMK = m <LMN - m < KMN

= 98 - 26

= 72 degrees.

Now triangle LKM is also isosceles since K is the center of the circle so

m < LKM = 180 - 2* 72

= 180 - 144

= 36 degrees.

4 0

3 years ago

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

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