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yarga [219]
3 years ago
7

Please help, im seriously stuck on this

Mathematics
1 answer:
Rasek [7]3 years ago
7 0

Answer:

x=64

Step-by-step explanation:

the shapes are similar

since the difference between 20 and 24 is 4

then you add 4 to 60 and get 64

I don't actually know 100% that this is correct but it makes sense in my head

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A right cylinder has a radius of 2 units and height of 5 units. What is the volume of the cylinder? Round to the nearest tenth.
Free_Kalibri [48]

Hi,

The volume of the cylinder is :

V = π × r^2 × h

V = 3,142 × (2)^2 × 5

V = 62, 8319 cm^3

π : Pi => 3,142

r : radius of the cylinder (2 units).

h : height of the cylinder (5 units).

The base of the cylinder is a circle , so to calculate the area we have to use the formula : area = π × r^2. And to find the volume , we have to multiply the area by the height.

•It was nice to help you, Bellalocc!

8 0
3 years ago
X 2 +10x+22=13x, squared, plus, 10, x, plus, 22, equals, 13 1) Rewrite the equation by completing the square. Your equation shou
Masteriza [31]

Answer:

(x + 5)^2 = (5 - 3)^2

Step-by-step explanation:

Given the equation;

x^2 + 10x + 22 = 13

We have;

x^2 + 10x = 13 - 22

x^2 + 10x = -9

Adding half of 10 to both sides and completing the square as usual, we have;

(x + 5)^2 = 5^2 + (-9)

(x + 5)^2 = (5 - 3)^2

5 0
3 years ago
Read 2 more answers
A salesperson sells $2200 in merchandise and earns a commission of $286.
Elza [17]

Answer:

13%

Step-by-step explanation:

Commission rate = commission / value sold

= 286/2200

= 13%

8 0
3 years ago
Read 2 more answers
How do you write four and 13 twentyeths as a decimal
deff fn [24]
4.65
13 ÷ 20 is 0.65, just add the 4
7 0
3 years ago
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Find the inverse of the matrix , if it exists
Montano1993 [528]

Answer:

So, option A is correct.

Step-by-step explanation:

To find the inverse of the matrix, the formula is:

A^{-1} = 1/|A| * Adj A

Inverse of the matrix exists if |A| does not equal to 0

|A| = 7 * 9 -( -2 * -10)

|A| = 63 - 20

|A| = 43

AS |A| ≠ 0

so inverse exist.

now finding Adj A

Adj A = \left[\begin{array}{cc}9&2\\10&7\end{array}\right]

\left[\begin{array}{cc}9/43&2/43\\10/43&7/43\end{array}\right]

So, option A is correct.

8 0
3 years ago
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