All categories

2 years ago

Solve the following inequality: |3x|>x+20

Mathematics

1 answer:

nikdorinn [45]2 years ago

7 0

Answer:

x>10,x<−5

Step-by-step explanation:

You might be interested in

Find the sum of two consecutive numbers if the smaller one is 43

Alexxx [7]

Answer:

Step-by-step explanation:

Smaller number = 43

Bigger number= 44

Sum = 43+44

8 0

2 years ago

Read 2 more answers

HELPPP only got 43 minutes to answer questions

irga5000 [103]

Answer:

20 feet below sea level

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

What does the suffix -ose likely indicate about a molecule?

STatiana [176]

The right answer is C

5 0

2 years ago

Angle A is 5x-15. Angle B is 2x+21. Solve for x then find the measure for Angle B

strojnjashka [21]

Answer:

×3-13 ×4-14 ×5-15 ×6-16

4 0

2 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

2 years ago