The area of a circle is 19.7cm². what is the radius?

Can anyone help?! Will mark brainliest!

IRINA_888 [86]

Answer:

Volume are over 120-177 because the layers are communicate each factors and angles..

Step-by-step explanation:

both 6 have 3 factors

A are for equation

5 centimeters are not allowed to communicate with other lessons and Numerator numbers...

HOPE IT HELPS!!

4 0

2 years ago

How to check division with multplication<br><img src="https://tex.z-dn.net/?f=3%20%5Ctimes%20292%20" id="TexFormula1" title="3 \

svetoff [14.1K]

Answer: By cross multiplication.

Step-by-step explanation: Given product is 3 × 292.

We know that after simple multiplication, we get 3 × 292 = 876.

Now, to check division with multiplication, either we need to divide 876 by 3 to get the answer 292,

or

we need to divide 876 by 292 to get the answer 3.

We will do that as follows -

$\dfrac{876}{3}=292~~~\textup{and}~~~\dfrac{876}{292}=3.$

Thus, doing cross-multiplication, we arrive at our conclusion.

6 0

3 years ago

An orange juice company sells a can of frozen orange juice that measures 9.4 centimeters in height and 5.2 centimeters in diamet

drek231 [11]

<h2>Hello!</h2>

The answer is:

The third option:

2.7 times as much.

To calculate how many more juice will the new can hold, we need to calculate the old can volume to the new can volume.

So, calculating we have:

Old can:

Since the cans have a right cylinder shape, we can calculate their volume using the following formula:

$Volume_{RightCylinder}=Volume_{Can}=\pi r^{2} h$

Where,

$r=radius=\frac{diameter}{2}\\h=height$

We are given the old can dimensions:

$radius=\frac{5.2cm}{2}=2.6cm\\\\height=9.4cm$

So, calculating the volume, we have:

$Volume_{Can}=\pi *2.6cm^{2} *9.4cm=199.7cm^{3}$

We have that the volume of the old can is:

$Volume_{Can}=199.7cm^{2}$

New can:

We are given the new can dimensions, the diameter is increased but the height is the same, so:

$radius=\frac{8.5cm}{2}=4.25cm\\\\height=9.4cm$

Calculating we have:

$Volume_{Can}=\pi *4.25cm^{2} *9.4cm=533.40cm^{3}$

Now, dividing the volume of the new can by the old can volume to know how many times more juice will the new can hold, we have:

$\frac{533.4cm^{3} }{199.7cm^{3}}=2.67=2.7$

Hence, we have that the new can hold 2.7 more juice than the old can, so, the answer is the third option:

2.7 times as much.

Have a nice day!

3 0

2 years ago

Find the area of the shape shown below.<br>

neonofarm [45]

Answer:

7

Step-by-step explanation:

First ignore the cut corner and pretend that it is just a rectangle.

The rectangle's area would be 2×4=8.

Then find out the cut corner's area.

The area of that triangle would be 1.

8-1=7 So the area of the shape is 7.

3 0

2 years ago

Read 2 more answers

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

1.

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

2.

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

2 years ago

The area of a circle is 19.7cm². what is the radius?​

The area of a circle is 19.7cm². what is the radius?