All categories

3 years ago

How who I complete the table?

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

Answer:

Step-by-step explanation:

you have to remove // to calculate

You might be interested in

What is the property of equality for AM = AM

Andrew [12]

Answer:

Reflexive

Step-by-step explanation:

When a=a, it is called the reflexive property

a=b then b=a is the symmetric property

a=b b=c then a=c is the transitive property

7 0

2 years ago

Please keep as simple as possible

Alenkinab [10]

Answer:

f(x) = (X^3 +6) + 4

Step-by-step explanation:

to move left add +6 from inside the parenthesis

to move up add +4 outside of parenthesis

*since there were no parenthesis you add them*

7 0

2 years ago

Read 2 more answers

Which of these options is not a quadratic equation in x?

il63 [147K]

Answer:

3x³ - 2x² + 1 = 0

Step-by-step explanation:

By definition, a quadratic equation cannot have an exponent higher than 2.

6 0

3 years ago

Read 2 more answers

) 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.

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