All categories

3 years ago

Prove that the two circles shown below are similar.

Mathematics

2 answers:

cestrela7 [59]3 years ago

6 0

Answer:

Well we need to find the area of circles B and D

We’ll use the following formula,

$\pi r^2$

The radius of the

is circle is 4 units,

so we plug 4 into $\pi r^2$ .

pi(4)^2

4*4 =16

16 * pi = 50.2654824574

The answer is 50 rounded to the nearest whole number.

The radius is 2,

pi(2)^2

2*2 = 4

pi * 4 = 12.5663706144

Or 13 Rounded to the nearest whole number.

So now we do 50 ÷ 13 = 3.84615384615

Which is about 4 rounded to the nearest whole number.

<em>The two circles are similar just that circle D is multiplied by a scale factor of 4 to get circle B</em>

jolli1 [7]3 years ago

5 0

Answer:

They have a scale factor of 2

They are both the same shape

The big circle has a radius of 4 and the small circle has a radius of 2. You divide the bigger circle by the smaller circle. 4/2=2

You might be interested in

Straight line depreciation. A car has an initial value of 29,564 and depreciates 532 a year . V represents value of car after t

LenaWriter [7]

V=29,564-532t would be the equation for that scenario

5 0

3 years ago

( - 37 + 37 ) ÷ - 15 = What is the answerr

Eduardwww [97]

Answer:

Step-by-step explanation:

$~~~(-37+37 )\div(-15)\\\\=0\div (-15)\\\\=0$

5 0

2 years ago

Read 2 more answers

What is the area of this rectangle? 10units 20units 24units 48 units

aksik [14]

Answer:

24 units ^2

Step-by-step explanation:

The base runs from 3 to 7 so the base is 7-3 = 4 units

The height runs from 2 to 8 so the height is 8-2 = 6 units

The area is

A = bh

= 6*4 = 24 units ^2

4 0

1 year ago

3. Simplify: 13.6 - 4.1 - 3.8 + 6.7

Art [367]

Answer:

12.4 is the answer

=13.6+6.7-4.1-3.8

=20.3-7.9

=12.4

5 0

2 years ago

Read 2 more answers

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago