All categories

3 years ago

Please help me please

Mathematics

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

first was the tan4cot

second was (-g,f)

You might be interested in

I hate quadratic formula

bezimeni [28]

Me too they suck and are hard

7 0

3 years ago

Read 2 more answers

Qual é o MMC entre 6 e 9???

Inessa05 [86]

Answer:

wut

Step-by-step explanation:

wut r u sayin

8 0

4 years ago

Am I on the right track?

alekssr [168]

if you want to be sure :/

7 0

4 years ago

Pleaseee help answer correctly !!!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!

Alexeev081 [22]

Answer:

108º

Step-by-step explanation:

to find the sum of interior angles: ( n − 2 ) × 180 = x

x = 540º

n = # of interior angles

first, solve for n

(n - 2) • 180 = 540 distribute to 180 to (n - 2)

180n - 360 = 540 add 360 to both sides

180n = 900 divide both sides by 180

n = 5

to find measure of each angle:

540 ÷ 5 = 108º

5 0

3 years ago

P=(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} ) : \frac{4x}{(x-1)^{2} }

balandron [24]

The solution to the division of the given surd is: $\mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x} }$

<h3>Division of Surds.</h3>

The division of surds follows a systemic approach whereby we divide the whole numbers separately and the root(s) are being divided by each other.

Given that:

$\mathbf{P=(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} ) : \frac{4x}{(x-1)^{2} }}$

i.e.

$\mathbf{=\dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )}{ \frac{4x}{(x-1)^{2} }} }$

Using the fraction rule:

$\mathbf{\dfrac{a}{\dfrac{b}{c}}= \dfrac{a\times c}{b}}$

$\mathbf{\implies \dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )(x-1)^{2}}{4x}} }$

By simplification, we have:

$\mathbf{ =\dfrac{\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{2} }{4x} }$

$\mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x} }$

Learn more about evaluating the division of surds here:

https://brainly.in/question/27942899

#SPJ1

6 0

2 years ago

Please help me please​

Please help me please