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2 years ago

CAN SOMEONE HELP ME WITH THIS PLZ! <3

Mathematics

2 answers:

Anestetic [448]2 years ago

7 0

The answer is D because to calculate circles area you multiply pi by the radius multiplied by two

coldgirl [10]2 years ago

6 0

Answer:

D. A = πr^2

Step-by-step explanation:

The formula of the area of a circle is <u>A = πr^2</u>.

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3x^2+7x+3<br> + 5x^2+12x

Dvinal [7]

Answer:

8x^2+19x

Step-by-step explanation:

Simplify the expression by combining like terms. Terms which look a like and sound a like should be added together.

3x^2+7x+3+5x^2+12x

x^2 terms: 3x^2+5x^2=8x^2

x terms: 7x+12x=19x

constants: 3

Together is is 8x^2+19x+3.

5 0

2 years ago

Read 2 more answers

Suppose n represents an even number for a simplified expression that represents the product of the next two even numbers

Digiron [165]

N+1 x n+2 = whatever that first part means

3 0

2 years ago

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

2 years ago

Read 2 more answers

Convert the angle \theta=\dfrac{8\pi}{9}θ=

Olegator [25]

Answer:

160°

Step-by-step explanation:

-Given the angle in radians as $\theta=\dfrac{8\pi}{9}\ rad$