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3 years ago

Multiply (2 4/5) (-5 2/3) A= -10 8/15 B=-15 13/15 C= 10 8/15 D= 15 13/15

Mathematics

1 answer:

icang [17]3 years ago

4 0

<h3>I'll teach you how to solve (2 4/5) (-5 2/3)</h3>

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(2 4/5) (-5 2/3)

Convert mixed numbers to improper fractions:

2 4/5 = 14/5

2 4/5 = 17/3

(14/5)(-17/3)

Remove parentheses:

-14/5 *-17/3

Multiply fractions:

14*17/5*3

Multiply:

-238/15

Convert improper fraction to mixed number:

-15 13/15

Your Answer Is B=-15 13/15

plz mark me as brainliest :)

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Answer:

(There's a confusion here, is x 2 x times 2? The answer and explanation are based on x multiply by 2)

y = -8

Step-by-step explanation:

Substitution Method-

y = x 2 - 4

x = -2

y = (-2) 2 - 4

PEMDAS - parentheses; exponents; Multiplication & Division; Addition & Subtraction

we are going to multipl -2 and 2 together ---> -4

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2 years ago

Solve for the indicated values.<br> For both

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CH = 22°

Step-by-step explanation:

We know that Arc AC and Arc CH both equal to 180°, therefore we can say;

AC + CH = 180°

158 + CH = 180°

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5 0

3 years ago

If we assume the population of Grand Rapids is growing at a rate of approximately 4% per decade, we can model the population fun

saw5 [17]

Answer:

The average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ] are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

Step-by-step explanation:

The given function is

$P(t)=181843(1.04)^{(\frac{t}{10})}$

where, P(t) is population after t years.

At t=5,

$P(5)=181843(1.04)^{(\frac{5}{10})}=185444.20$

At t=6,

$P(6)=181843(1.04)^{(\frac{6}{10})}=186172.95$

At t=7,

$P(7)=181843(1.04)^{(\frac{7}{10})}=186904.57$

At t=8,

$P(8)=181843(1.04)^{(\frac{8}{10})}=187639.06$

At t=9,

$P(9)=181843(1.04)^{(\frac{9}{10})}=188376.44$

At t=10,

$P(10)=181843(1.04)^{(\frac{10}{10})}=189116.72$

The rate of change of P(t) on the interval $[x_1,x_2]$ is

$m=\frac{P(x_2)-P(x_1)}{x_2-x_1}$

Using the above formula, the average rate of change of the population on the intervals [ 5 , 10 ] is

$m=\frac{P(10)-P(5)}{10-5}=\frac{189116.72-185444.20}{5}=734.504$

The average rate of change of the population on the intervals [ 5 , 9 ] is

$m=\frac{P(9)-P(5)}{9-5}=\frac{188376.44-185444.20}{4}=733.06$

The average rate of change of the population on the intervals [ 5 , 8 ] is

$m=\frac{P(8)-P(5)}{8-5}=\frac{187639.06-185444.20}{3}=731.62$

The average rate of change of the population on the intervals [ 5 , 7 ] is

$m=\frac{P(7)-P(5)}{7-5}=\frac{186904.57-185444.20}{2}=730.185$

The average rate of change of the population on the intervals [ 5 , 6 ] is

$m=\frac{P(6)-P(5)}{6-5}=\frac{186172.95-185444.20}{1}=728.75$

Therefore the average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ] are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

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4 years ago

The graph of f(x) = |xl is transformed to g(x) = |x+1|- 7. On which interval is the function decreasing?

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Answer:c

Step-by-step explanation:

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3 years ago

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mr_godi [17]

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3 years ago

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