3 years ago

WILL MARK BRAINLIEST AND 5 STARS!!!!!

Mathematics

1 answer:

Afina-wow [57]3 years ago

7 0

1) 2/1

2) 0

3)1/2

............

You might be interested in

Where decimal go in 288

Andreyy89

Be more specific. The decimal can be anywhere give more detail

6 0

3 years ago

Calculate the mean of this data set: 12, 35, 44, 74, 23, 49, 45, 18, 90, 56, 84,

Blababa [14]

Well, here is the mean of ur data collected 48.181818...

4 0

3 years ago

Read 2 more answers

What is a triangle equal to

VashaNatasha [74]

Answer:

A triangle equals 180 degrees.

4 0

3 years ago

Read 2 more answers

What is the result of adding these two equations? \begin{aligned} 2x+3y &= -5 \\\\ 5x-y &= -12 \end{aligned} 2x+3y 5x−y

EleoNora [17]

Answer:

$7x+2y = -17$

Step-by-step explanation:

We have the following equations:

$2x+3y = -5$

$5x-y = -12$

If we want to add these equations, we just need to add x-terms, y-terms and independent terms.

$(2x+5x)+(3y-y) = -5-12$

$7x+2y = -17$

I hope it helps you!

7 0

3 years ago

Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x

svetlana [45]

Hello,

$sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\

$
==============================================================
$2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 
$

8 0

3 years ago

Read 2 more answers