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Alexxx [7]
3 years ago
15

What value of x will make the two triangles similar? plz help

Mathematics
1 answer:
MrRa [10]3 years ago
7 0

Answer:

x = 60

Step-by-step explanation:

The Factor of sides is 1.5 and I found that out by dividing one side to the identical side on the other triangle then you multiply the bottom side by 1.5 to get the answer

please give brainliest

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MAJOR HELP!
Feliz [49]

Answer:

The radius of the cone is 21 units

The volume of the cone is 9702 units³

Step-by-step explanation:

* Lets revise the area of the circle and the volume of the cone

- The base if the cone is a circle

- The area of the circle is πr², where r is the radius of the circle

∵ The area of the base of the cone = 1386 units²

∵ The area of the base = πr²

∵ π = 22/7

∴ 1386 = 22/7 (r²) ⇒ multiply each side by 7

∴ 9702 = 22 r² ⇒ divide both sides by 22

∴ 441 = r² ⇒ take √ for both sides

∴ 21 = r

* The radius of the cone is 21 units

- The volume of the cone = 1/3 πr²h , where h is the height of the cone

∵ The cone's height is equal to the radius of its base

∴ h = r

∵ The volume of the cone = 1/3 πr²h

∴ The volume of the cone = 1/3 πr²(r) = 1/3 πr³

∵ r = 21 units

∴ The volume of the cone = 1/3 (22/7) (21)³

∴ The volume of the cone = 9702 unit³

* The volume of the cone is 9702 units³

3 0
4 years ago
Blank thousands = 1800 tens
Alexxx [7]
 the answer is: 18 thousands = 1800
6 0
4 years ago
Read 2 more answers
Helppp it’s due in a couple minutes!!!
tiny-mole [99]
Not my work. but found a answer key to what u need.

3 0
2 years ago
Is y=x/10 represent proportional relationship
larisa86 [58]

Yes, because x turns to into 1/10.

5 0
3 years ago
For the matrices below. Find all (real) eigenvalues. <br><br> A= [7 9]<br> [0 9]
Jet001 [13]

Answer:

The answer is "9 and 7".

Step-by-step explanation:

Given:

A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]

Using formula:

|A-\lambda \cdot I|= 0\\\\

\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\

\to \lambda=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda=9\\\\

8 0
3 years ago
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