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SCORPION-xisa [38]
3 years ago
15

??????????????????????

Mathematics
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

254.34

Step-by-step explanation:

Good luck!

irga5000 [103]3 years ago
4 0

Answer:

A = 254.34

Step-by-step explanation:

Area of a circle formula:

A = πr²

r = d/2 = 9

A = 3.14(9²)

A = 3.14(81)

A = 254.34

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Factor the expression. 4x(x^3 − 6) − 7(x^3 − 6) HELP ASAP!!
NikAS [45]
4x(x^3 - 6) - 7(x^3 - 6)

<em>Take out the common factor,  (x³ - 6), from both terms:</em>
= (x^3 - 6)(4x - 7) 

\Longrightarrow Answer \ D

8 0
3 years ago
Is the number 3 multiple of 9
defon
No

Imagine multiples to be 'times tables'. So the multiples of 9 are the 9 times table:

9, 18, 27, 36 and so on.

3 would be a factor of 9. Factors are numbers that can divide exactly into another number.

I hope I helped!
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3 years ago
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6 0
2 years ago
Given the quadratic equation 8-6x=-2x^2 complete the following:
Ipatiy [6.2K]

Answer:

a. 2x^2-6x+8

b. a=2, b=-6, and c=8

Step-by-step explanation:

The standard form of a quadratic is ax^2+bx+c written in descending order and can be found by using inverse operations to rearrange terms.

8-6x=-2x^2\\\\8-6x+2x^2=-2x^2+2x^2\\\\2x^2-6x+8=0

8 0
3 years ago
How do you prove cosx = 1-tan^2(x/2)/1+tan^2(x/2)?
nikdorinn [45]
\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=&#10;\begin{cases}&#10;\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}&#10;\\ \quad \\&#10;\boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}}&#10;\\ \quad \\&#10;\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}&#10;\end{cases}\\\\&#10;-------------------------------\\\\&#10;tan^2\left( \frac{x}{2} \right)\implies \left[ \cfrac{sin(x)}{1+cos(x)} \right]^2\implies \cfrac{sin^2(x)}{[1+cos(x)]^2}&#10;\\\\\\&#10;\boxed{\cfrac{sin^2(x)}{1+2cos(x)+cos^2(x)}}

now, let's plug that in the right-hand-side expression,

\bf cos(x)=\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\\\\&#10;-------------------------------\\\\&#10;\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\implies \cfrac{1-\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}{1+\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}&#10;\\\\\\&#10;\cfrac{\frac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)}}{\frac{1+2cos(x)+cos^2(x)~+~sin^2(x)}{1+2cos(x)+cos^2(x)}}

\bf \cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{\underline{1+2cos(x)+cos^2(x)}}\cdot \cfrac{\underline{1+2cos(x)+cos^2(x)}}{1+2cos(x)+cos^2(x)~+~sin^2(x)}&#10;\\\\\\&#10;\cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)~+~sin^2(x)}

\bf -------------------------------\\\\&#10;recall\qquad sin^2(\theta)+cos^2(\theta)=1\\\\&#10;-------------------------------\\\\&#10;\cfrac{\boxed{sin^2(x)+cos^2(x)}+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+\boxed{1}}&#10;\\\\\\&#10;\cfrac{cos^2(x)+2cos(x)+cos^2(x)}{2+2cos(x)}\implies \cfrac{2cos(x)+2cos^2(x)}{2+2cos(x)}&#10;\\\\\\&#10;\cfrac{\underline{2} cos(x)~\underline{[1+cos(x)]}}{\underline{2}~\underline{[1+cos(x)]}}\implies cos(x)
8 0
3 years ago
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