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3 years ago

Mentioneaza rolul indicatiei scenei : vrea sa para nepasator

Mathematics

1 answer:

melomori [17]3 years ago

7 0

Answer:

Step-by-step explanation:

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paul has a total of 15 markers and pencil. Each pencil cost $0.30 and each marker costs $1.25 how many pencils did paco buy.

Mariana [72]

Answer: We need how much he spent on the whole thing

Step-by-step explanation:

It isn't really specific because he could of bought 6 pencils and 9 markers for all we know so we need the total how much he spent on all 15 markers and pencils

5 0

3 years ago

Pls help (50 points)

svetlana [45]

Answer:

4.5 inches

Step-by-step explanation:

We are given the formula:

V = $\pi r^{2}h$

Plug in the the radius (3) and volume (127.23):

127.23 = $3.14(3)^{2} h$

Square the 3 :

127.23 = $3.14(9)h$

Multiply the pi & volume:

127.23 = 28.26h

Divide both sides by 28.26:

127.23 ÷ 28.26 = 28.26h ÷ 28.26

h = 4.5021231...

I hope this helps!

6 0

3 years ago

Read 2 more answers

janis and susan went to a clearance sale at a clothing store all blouses were selling for the same price and all skirts were sel

alexandr402 [8]

$8.5, $\frac{68}{8} = 8.5$

5 0

3 years ago

A tin is a right circular cylinder with a diameter of 6 meters and a height of 10 meters. What is the surface area of this tin,

Alex777 [14]

Answer:

245.04

Step-by-step explanation:

A=2πrh+2πr2=2·π·3·10+2·π·32≈245.04423

Divide 6 by 2 to find the radius

5 0

3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

3 years ago