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Semenov [28]
3 years ago
12

The radius of a cylinder is 3x – 2 cm. The height of the cylinder is x +3 cm. What is the

Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0

Surface Area of the cylinder is 6x² + 83x - 58 cm²

Step-by-step explanation:

  • Step 1: Given surface area of the cylinder, A = 23r + 2rh. Here, r = (3x - 2) and h = (x + 3)

⇒ A = 23(3x - 2) + 2(3x - 2)(x + 3)

⇒ A = 69x - 46 + (6x² + 18x - 4x - 12)

⇒ A = 6x² + 83x - 58 cm²

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Will mark u brainliest <br> 15 points!!!!<br> Class 9 math
KonstantinChe [14]

Answer:

\frac{3375}{512} or ≈6.59

Step-by-step explanation:

We can first begin by simplifying each of the numbers with exponents. Recall that in a fraction exponent, the numerator is the power, while the denominator is the root.

Take 25\frac{3}{2} for example. The '2' in the fraction means we must take the square of 25. √25 = 5.

The '3' in the fraction means we take the power, which means we must cube '5'.

5³ = 125. Therefore, 25\frac{3}{2}  = 125. Use this process for the other numbers:

25^{\frac{3}{2} } = 125\\243^{\frac{3}{5} } = 27\\16^{\frac{5}{4} } = 32\\ 8^{\frac{4}{3} } = 16

The new fraction would look like:

\frac{125 * 27}{32 * 16}

Which simplifies to:

\frac{3375}{512} or ≈6.59

8 0
3 years ago
Read 2 more answers
The graph of y = f (x) is given in the figure. Sketch = 1/2 ( + 2) − 2.
jolli1 [7]

Answer:

y=1/2+2

Step-by-step explanation:

8 0
3 years ago
Find y.<br> Do not round your answer.<br> 9 in<br> у<br> 5 in<br> Х<br> 19 in<br> y = [ ? ] in<br> =
Verizon [17]

Answer:

33y

Step-by-step explanation:

step 1. 9y+5y+9y=33y

7 0
2 years ago
Which statement about sqrt x-5 minus sqrt x=5 true?
Allushta [10]

Answer:

9 is an extraneous solution.

Step-by-step explanation:

Extraneous solution is a solution that when plugged into the equation do not hold true.

Here we are given an algebraic equation in terms of single variable x as:

\sqrt{x-5}-\sqrt{x}=5-----(1)

Now, we will solve this equation to obtain the solution as:

on squaring both side of the equation we obtain:

(\sqrt{x-5}-\sqrt{x})^2=5^2

x-5+x-2\sqrt{x-5}\sqrt{x}=25\\\\2x-5-2\sqrt{x-5}\sqrt{x}=25\\\\2x-30=2\sqrt{x-5}\sqrt{x}\\\\on\ dividing\ both\ side\ by\ 2\ we\ obtain:\\\\x-15=\sqrt{x-5}\sqrt{x}

Again on squaring both side of the equation we obtain:

x^2+225-30x=(x-5)x\\\\\x^2+225-30x=x^2-5x\\\\225=-5x+30x\\\\225=25x\\\\x=9

Now when we plug x=9 back to the original equation i.e. equation (1) we get:

\sqrt{9-5}-\sqrt{9}=5\\\\\sqrt{4}-\sqrt{9}=5\\\\2-3=5\\\\-1=5

Hence, the equation does not hold true.

Hence, 9 is a extraneous solution

5 0
3 years ago
Solve the system of linear equations. separate the x- and y- values with a coma. 20x=-58-2y
lianna [129]
The best way to solve is by using elimination method.
20x = -58 - 2y
17x = -49 - 2y
Multiply second equation by -1
20x = -58 - 2y
-17x = 49 + 2y
Add equations.
3x = -9 
Divide.
x = -3
Plug in -3 into one of the equations.
17(-3) = -49 - 2y
-51 = -49 - 2y
Add 49 to both sides.
-2 = -2y
Divide.
1 = y
So your solution is (-3, 1).
I hope this helps love! :)

6 0
3 years ago
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