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3 years ago

1 1/7 x 3/4 write in simplest form

Mathematics

1 answer:

Helen [10]3 years ago

8 0

Answer:

8/7 x 3/4 = 24/28= 6/7

Step-by-step explanation:

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What 2 ratio is equivalent to 4/14 HELPPPP ILL GIVE U 15 AND MARK BRAINLIST

sineoko [7]

Answer:

4:14 2:7 8:28

Step-by-step explanation:

7 0

3 years ago

Whats the distance between (-9,9) and (-4,4)

swat32

Answer:

I think the answer is (-5,5)

Step-by-step explanation:

Im not for sure so im sorry if it wrong

4 0

2 years ago

One positive number is 3 more than twice another. If their product is 629, find the numbers.

Soloha48 [4]

Answer:

17,37

Step-by-step explanation:

one number = x

The positive number = 2x + 3

x * (2x+3) = 629

x*2x + x *3 = 629

2x² + 3x - 629 = 0

2x² - 34x + 37x - 17*37 =0

2x*(x -17) + 37(x - 17) = 0

(x - 17)(2x + 37) = 0

x - 17 = 0 ; Ignore 2x + 37 as s is a positive number

x = 17

Positive number = 2*17 +3 = 34+3 = 37

5 0

3 years ago

Could someone explain how to do this question?

meriva

Answer:

$p\left( A|B\right) =0$

Step-by-step explanation:

Formula:

$p\left( A|B\right) =\frac{p\left( A\cap B\right) }{p\left( B\right) }$

The only prime outcome is 3

Then

The set A = {3}

1 and 2 are the divisors of 2

Then

The set B = {1 , 2}

Therefore , p(A∩B) = 0

Also p(B) = 0.4+0.1 = 0.5

$\Longrightarrow p\left( A|B\right) =\frac{0 }{0.5 }=0$

7 0

2 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago