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MA_775_DIABLO [31]
3 years ago
14

Circle with radius 5 inches. round your answer to the nearest tenth

Mathematics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

Perimeter = 31.4 inches

Area = 78.5 inches squared

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Please help, thank you :)
wolverine [178]

Answer:

y^6

Step-by-step explanation:

\frac{y}{1} \times   \frac{{y}^{5} }{1}

=

{y}^{6}

7 0
2 years ago
Write an algebraic expression for the following
MrRa [10]

Answer:

12t = c

Step-by-step explanation:

I believe that's it

8 0
2 years ago
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Work out the surface area of a cylinder when the height = 18cm and the volume = 1715cm cubed
lara [203]

Answer:

813.4 cm² (nearest tenth)

Step-by-step explanation:

<u>Volume of a cylinder</u>

\sf V=\pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

Given:

  • h = 18cm
  • V = 1715 cm³

Use the Volume of a Cylinder formula and the given values to find the <u>radius of the cylinder</u>:

\implies \sf 1715=\pi r^2 (18)

\implies \sf r^2=\dfrac{1715}{18 \pi}

\implies \sf r=\sqrt{\dfrac{1715}{18 \pi}

<u>Surface Area of a Cylinder</u>

\sf SA=2 \pi r^2 + 2 \pi r h \quad\textsf{(where r is the radius and h is the height)}

Substitute the given value of h and the found value of r into the formula and solve for SA:

\implies \sf SA=2 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)^2 + 2 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)(18)

\implies \sf SA=2 \pi \left(\dfrac{1715}{18 \pi} \right) + 36 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)

\implies \sf SA=\dfrac{1715}{9} + 36 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)

\implies \sf SA=813.3908956...

Therefore, the surface area of the cylinder is 813.4 cm² (nearest tenth)

8 0
2 years ago
Read 2 more answers
For each system of equations, drag the true statement about its solution set to the box under the system?
natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

 of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

   coefficients of x and y are the same and the numerical terms

   are the same

- The system of equation has one solution if at least one of the

  coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

  numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0
3 years ago
Find the value of x that make m || n
pickupchik [31]
The angle has to be a right angle so...
16x-6=90
16x=96
x=6
4 0
3 years ago
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