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melisa1 [442]
3 years ago
7

How who I complete the table?

Mathematics
1 answer:
Natalka [10]3 years ago
5 0

Answer:

6

7

8

10

Step-by-step explanation:

you have to remove // to calculate

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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 <u>inside</u> the parenthesis

to move up add +4 <u>outside</u> of parenthesis

<em>*since there were no parenthesis you add them*</em>

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}

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
What is the product of 1 2/3 and −3 1/2?
Fynjy0 [20]
<span>1 2/3 x  −3 1/2
= 5/3 x -7/2
= - 35/6
= - 5 5/6

answer
A. </span><span>−5 5/6 ​</span>
6 0
3 years ago
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