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ss7ja [257]
2 years ago
12

Divide please help I will mark as brainliest

Mathematics
2 answers:
noname [10]2 years ago
6 0

Answer:

<u>16</u>

Step-by-step explanation:

Dividing 2 fractions is the same as multiplying the fraction in the numerator with the inverse of the fraction in the denominator.

Therefore,

  • \frac{-20}{3} ÷ \frac{-5}{12}
  • -\frac{20}{3} × -\frac{12}{5}
  • Note : -ve signs cancel out, cross multiply 20 and 5 to get 4, and cross multiply 12 and 3 to get 4
  • ⇒ 4 x 4
  • ⇒ <u>16</u>
-BARSIC- [3]2 years ago
5 0

Answer: Heyaa!

Your Answer Is.. 16

Step-by-step explanation:

A.

Hopefully this helps <em>you!</em>

- Matthew

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The difference of the quotient m and 9 and the quotient of n and 30
Zanzabum
Answer: (30m-9n)/270

First we start with m/9-n/30.

We multiply the denominators to get 270.

Then we multiply the numerators by the original denominators to get (30m-9n).

To check, we can use m=2, and n=4

2/9-4/30=4/45

(30*2-9*4)/270=24/270=4/45
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3 years ago
Find the indicated function value for csc -450°
Mekhanik [1.2K]

Step-by-step explanation:

\csc( - 450 \degree) \\  \\  =  -  \csc450 \degree \\  \\  =  -  \csc(360 \degree + 90\degree) \\  \\  = -  \csc( 90\degree) \\  \\  =  - 1

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3 years ago
Will give brainliest! Probability
alex41 [277]

Answer:

1/49

Step-by-step explanation:

You wrote the problem correctly.  You just needed to multiply instead of add.

1/7*1/7=1/49

So the answer is 1/49.

3 0
3 years ago
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Find the prime factors.<br> 7) 48
FinnZ [79.3K]

Step-by-step explanation:

48 = 48 \times 1 \\ 48 = (24 \times 2) \times 1 \\ 48 = (12 \times 2) \times 2 \times 1 \\ 48 = (6 \times 2) \times 2 \times 2 \times 1 \\ 48 = (3 \times 2) \times 2 \times 2 \times 2 \times 1

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2 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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