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3 years ago

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

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!

3 0

3 years ago

X/7 > -2 <br>What is the answer and it says your answer must be simplified

VladimirAG [237]

Answer:

x>-14

Step-by-step explanation:

x/7>-2

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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]

$\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}
\end{cases}\\\\
-------------------------------\\\\
tan^2\left( \frac{x}{2} \right)\implies \left[ \cfrac{sin(x)}{1+cos(x)} \right]^2\implies \cfrac{sin^2(x)}{[1+cos(x)]^2}
\\\\\\
\boxed{\cfrac{sin^2(x)}{1+2cos(x)+cos^2(x)}}$

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

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

$\bf \cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{\underline{1+2cos(x)+cos^2(x)}}\cdot \cfrac{\underline{1+2cos(x)+cos^2(x)}}{1+2cos(x)+cos^2(x)~+~sin^2(x)}
\\\\\\
\cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)~+~sin^2(x)}$

$\bf -------------------------------\\\\
recall\qquad sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\
\cfrac{\boxed{sin^2(x)+cos^2(x)}+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+\boxed{1}}
\\\\\\
\cfrac{cos^2(x)+2cos(x)+cos^2(x)}{2+2cos(x)}\implies \cfrac{2cos(x)+2cos^2(x)}{2+2cos(x)}
\\\\\\
\cfrac{\underline{2} cos(x)~\underline{[1+cos(x)]}}{\underline{2}~\underline{[1+cos(x)]}}\implies cos(x)$

8 0

3 years ago