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QveST [7]
3 years ago
11

Jayne says that 0.45 is equal to 4/10. Is she correct? Explain.

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0
0.45 =  \dfrac{45}{100}

\dfrac{4}{10} =  \dfrac{4 \times 10}{10 \times 10} =  \dfrac{40}{100}

\dfrac{45}{100}  \neq  \dfrac{40}{100}

They are not the equal.
yKpoI14uk [10]3 years ago
4 0
No, 4/10 is equal to 0.40 and not 0.45 since 4/10 goes in tens.
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Solve 4x^2 - 36x + 32 = 0<br><br>Sorry! It's suppose to be + 32.<br>Please show all work.
mylen [45]

Answer:

x=1, x=8

Step-by-step explanation:

4x2−36x+32=0

Step 1: Factor left side of equation.

4(x−1)(x−8)=0

Step 2: Set factors equal to 0.

x−1=0 or x−8=0

 (+1)        (+8)

x=1 or x=8




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3 years ago
Make a frequency distribution and find the relative frequencies for the following number set. Round the relative frequency to th
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Number......frequency.....relative frequency
10                  2                    2/24 = 8.3%
20                  2                    2/24 = 8.3%
30                  4                    4/24 = 16.7%
40                  2                    2/24 = 8.3%
50                  3                    3/24 = 12.5%
60                  5                    5/24 = 20.8%
70                  3                    3/24 = 12.5%
80                  1                    1/24 = 4.2%
90                  2                    2/24 = 8.3%
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3 years ago
Write a real world situation that would representby the system y=3x +20 and y=5x + 10
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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}

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