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Lesechka [4]
3 years ago
11

In the poportion 1/z =4/5/8 which number is equal to z in the proportion

Mathematics
1 answer:
Nady [450]3 years ago
8 0

Answer:

case 1) z=10

case 2) z=5/32

Step-by-step explanation:

case 1) we know that

Using proportion

\frac{1}{z}=\frac{(4/5)}{8}

solve for z

\frac{1}{z}=\frac{(4/5)}{8}\\ \\z(4/5)=8\\ \\z=8*5/4\\ \\z=10

case 2) we know that

Using proportion

\frac{1}{z}=\frac{4}{5/8}

solve for z

\frac{1}{z}=\frac{4}{5/8}\\ \\4z=5/8\\ \\z=5/32

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Find the measures of a, b, c, and d.​
iris [78.8K]

Answer:

Step-by-step explanation:

a=55  (180-125)

b= 65 (180-55-60)

d= 30    (90-60)

 c=  60 (180-30-90)

8 0
3 years ago
A researcher compared the heights and shoe sizes for 50 men, selected at random. The equation shown describes a line of best hit
tresset_1 [31]

Answer:

7 1/2

Step-by-step explanation:

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6 0
3 years ago
If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

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

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

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

We know that:

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

This gives:

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

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

6 0
3 years ago
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maxonik [38]
C because i did it and got it right
4 0
3 years ago
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Finger [1]

Answer:

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Step-by-step explanation:

i just wrote it out and i think thats the answer but not sure. hopefully this helps :)

8 0
3 years ago
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