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3 years ago

What is equivalent to 6/8

Mathematics

2 answers:

Liono4ka [1.6K]3 years ago

3 0

3/4 is the reduced form, also 12/16 is each doubled

Leni [432]3 years ago

3 0

Reduce 6/8 by 2 to get 3/4. 3/4 is equal to 6/8.

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The area of a rectangle is 56 cm. The length is 2cm more than x and width is 5 cm less than twice x. Solve for x. Round to the n

ra1l [238]

Area = Length * Width = l * w
l = x + 2
w = 2x - 5
Area = (x + 2)(2x - 5) = 56
Area = 2x^2 -x - 10 = 56
Area = 2x^2 - x - 66 = 0 Now you have to factor this thing.
Area = (2x + 11)( x - 6) = 0 This is more or less done by guessing.
2x + 11 = 0 can't work. x would be negative.
x - 6 = 0
x = 6.

Check
=====
l = x + 2
l = 6 + 2
l = 8

w = 2*x - 5
w = 2*6 - 5
w = 12 - 5
w = 7

l*w = 8*7 = 56. It checks.
If this is the best answer you get, could you consider a Brainly? I need 2 of them to move up.

3 0

3 years ago

Solve 5x=125. X=25 x=3

Zolol [24]

Answer: X = 25

Step-by-step explanation: 5 * 25 = 125

5 0

3 years ago

Please Help! Please! I don’t understand!

Brums [2.3K]

Answer:

#1 false

#2 true

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Solve for x.

denis-greek [22]

Answer:

$\frac{3\±\sqrt{3}}{3}$

Step-by-step explanation:

Whe have the function $3x^2-6x+2=0$

To solve this equation we must factor.

To factor, we need to find the points where the function is equal to zero. So we use the quadratic formula

$\frac{-b\±\sqrt{b^2-4ac}}{2a}$

Where:

a = 3

b = -6

c = 2

Then

$\frac{-(-6) + \sqrt{(-6)^2-4(3)(2)}}{2(3)} = \frac{3+\sqrt{3}}{3}\\\\and\\\\\frac{-(-6) - \sqrt{(-6)^2-4(3)(2)}}{2(3)}= \frac{3-\sqrt{3}}{3}$

finally the correct answer is $\frac{3\±\sqrt{3}}{3}$

5 0

3 years ago

Read 2 more answers

Question in picture A) 16/63 B)-16/63 C) 63/16 D) -63/16

Orlov [11]

ANSWER
$\tan(x + y) = - \frac{63}{16}$

EXPLANATION

We were given that,

$\csc(x) = \frac{5}{3}$

This implies that,

$\sin(x) = \frac{3}{5}$

We use the Pythagorean identity

$\sin^{2} (x) + \cos^{2} (x)= 1$
to get,

$\cos(x) = \sqrt{1 - ( { \frac{3}{5} })^{2}} = \frac{4}{5}$

We were also given that,

$\cos(y) = \frac{5}{13}$

This means that,

$\sin(y) = \sqrt{1 - {( \frac{5}{13}) }^{2} } = \frac{12}{13}$

This is because,

$0 < \: x \: < \frac{\pi}{2}$

$0 < \: y \: < \frac{\pi}{2}$

This angles are in the first quadrant so we pick the positive values.

$\tan(x + y) = \frac{ \sin(x + y) }{ \cos(x + y) }$

$\tan(x + y) = \frac{ \sin(x ) \cos(y) + \sin(y) \cos(x) }{ \cos(x) \cos(y) - \sin(x) \sin(y) }$

$\tan(x + y) = \frac{ \frac{3}{5} \times \frac{5}{13} + \frac{12}{13} \times \frac{4}{5} }{ \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13} }$

$\tan(x + y) = - \frac{63}{16}$

The correct answer is D

4 0

3 years ago