All categories

3 years ago

What is 25/30 simplified(25/30 is a fraction)

Mathematics

2 answers:

Irina-Kira [14]3 years ago

8 0

It's 4/3 poiojehnemejsiiekeje

dybincka [34]3 years ago

7 0

You first have to find a multiple that can divide both numbers equally. 25 divided by 5 is 5. And 30 divided by 5 equals 6. So the simplified fraction of 25/30 is 5/6

You might be interested in

What is 50g/cm= kg/m

Anni [7]

The answer is 5 kg/m

6 0

3 years ago

Read 2 more answers

What is 0.7 divided by 792

Dvinal [7]

0.7/729 equals 0.00088383838

6 0

3 years ago

Read 2 more answers

Perfect square between 1 and 9

Rasek [7]

Answer:

Step-by-step explanation:

4 0

3 years ago

If f(x) = 3^x + 10x and g(x) = 2x - 4, find (f + g)(x)

kow [346]

Answer:

The Answer is D

Step-by-step explanation:

All you do is add both equations 3^x+10x+2x-4

then you add the like varibles which is 10x+2x=12x

So, the answer is 3^x+12x-4

6 0

3 years ago

Read 2 more answers

Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

2 years ago