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3 years ago

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For every 40 phone calls that Jess made in a month, she received 70 phone calls. What is the ratio in simplest form of the numbe

r of phone calls made to the number of phone calls received by Jess that month?

2 answers:

Anit [1.1K]3 years ago

5 0

Answer:

simplified answer is 4:7

insens350 [35]3 years ago

4 0

The simplified answer is 4:7

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A company has two packaging machines in a unit, each with a different daily capacity. The capacity of machine 1 is defined by th

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Total capacity = sum of the individual production capacities.

Here,

Total capacity = sum of f(m) = (m + 4)^2 + 100 and g(m) = (m + 12)^2 − 50.

Then f(m) + g(m) = (m + 4)^2 + 100 + (m + 12)^2 − 50.

We must expand the binomial squares in order to combine like terms:

m^2 + 8 m + 16 + 100

+m^2 + 24m + 144 - 50

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Then f(m) + g(m) = 2m^2 + 32m + 160 + 50

f(m) + g(m) = 2m^2 + 32m + 210, where m is the number of

minutes during which the two machines operate.

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3 years ago

How to find the measure of the angle is you have the incenter?

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mark is drawing a map of his favorite park. the park is 3 miles long and 1.5 miles wide. the map scale is 1 inch = .25 miles. ho

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1.5 miles wide * 1 inch / 0.25 miles = 6 inches :D

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Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

so

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

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Leya [2.2K]

Answer:

3.16

Step-by-step explanation:

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