3 years ago

How to solve 2 3/4 - 1 5/8

Mathematics

1 answer:

daser333 [38]3 years ago

6 0

Answer:

1 1/8

Step-by-step explanation:

First, Make the first fraction into a new one with an equal denominator. (3/4 x2= 6/8)

Then, subtract the whole numbers (2-1=1) and the fractions (6/8-5/8= 1/8)

Lastly, add your two differences together and you get 1 1/8.

You might be interested in

Three muffin pans with six in each pan what is the fraction for this

irakobra [83]

Answer:

1/18 ?

Step-by-step explanation:

I'm not sure if you're written the whole question?

3 pans with 6 muffins in each pan, so total number of muffins = 3 x 6 = 18

So each muffin is 1/18

4 0

3 years ago

Read 2 more answers

What addition doubles fact can help you find 4+3

PtichkaEL [24]

The answer would be 7

8 0

3 years ago

Solve for y 3x+11y=-5

mars1129 [50]

$\displaystyle 3x+11y=-5\quad\Big| -3x\\\\11y=-5-3x\quad\Big|:11\\\\y= -\frac{5+3x}{11}$

6 0

3 years ago

What is the quadratic function that is created with roots at 2 and 4 and a vertex at (3, 1)?

Arada [10]

Hello,

y=k*(x-2)(x-4)
and is passing throught (3,1)
==>1=k*(3-2)(3-4)==>k=-1

y=-(x-2)(x-4) is an answer

4 0

3 years ago

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

8 0

3 years ago