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3 years ago

What is the volume of a cylinder with a radius of 5 cm and height of 10 cm? In terms of pi.

Mathematics

1 answer:

natka813 [3]3 years ago

5 0

Answer:

Volume of the cylinder=250π cm³

Step-by-step explanation:

The formula for calculating the volume of the cylinder is;

V=πr²h

where;

V=volume of the cylinder

π=pi

r=radius of the cylinder

h=height of the cylinder

In our case;

V=unknown

r=5 cm

h=10 cm

replacing;

V=π×5²×10=250π cm³

Volume of the cylinder=250π cm³

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2ax-b=cx+d solve for x

Alexxandr [17]

Answer:

x = (d + b) / (2a - c)

Step-by-step explanation:

2ax-b = cx+d

2ax - cx = d + b

x (2a-c) = d + b

x = (d + b) / (2a - c)

7 0

3 years ago

Need help simple explanation if possible

photoshop1234 [79]

Answer:

m < A = 60º

m < B = 30º

Step-by-step explanation:

The given sides on this triangle are: 6, $6\sqrt{3}$ , and 12

Any triangle with the angles of 30º - 60º - 90º always has side lengths in this proportion:

x, $x\sqrt{3}$ , 2x

We can line this up with the given sides. If x is 6, then 2x would be 12.

x : $x\sqrt{3}$ : 2x = 6 : $6\sqrt{3}$ : 12

Angle B is across from 6, the shortest side. That also means that it corresponds to x, or the smallest angle in the proportion, 30º.

m < B = 30º

Solving for < A:

Method 1) Sum of Angles in a Triangle

Since we already know that one angle is right and therefore 90º and m < B is 30º, we can subtract these from the total sum of angle measures in a triangle to get the last angle, < A.

180º - 90º - 30º = 60º

m < A = 60º

Method 2) Using the second part of the proportion

Since m < A is across from the second largest side, we know that it is equal to $x\sqrt{3}$ ( $6\sqrt{3}$ in this question) or 60º in the angle proportion.

This means that m < A = 60º

Let me know if you have any questions!

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2 years ago

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2 years ago

7x-3y=-6 convert to slope intercept form

r-ruslan [8.4K]

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4 years ago

2 Here are two equations:

MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

i (3,4)

Putting in Equation 1:

$6(3) + 4(4) = 34\\18+16=34\\34=34\\$

Putting in Equation 2:

$5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15$

ii. (4,2.5)

Putting in Equation 1:

$6(4) + 4(2.5) = 34\\24+10=34\\34=34\\$

Putting in Equation 2:

$5(4) - 2(2.5) = 15\\20-5 = 15\\15=15$

ill. (5,5)

$6(5) + 4(5) = 34\\30+20=34\\50\neq 34$

Putting in Equation 2:

$5(5) - 2(5) = 15\\25-10 = 15\\15=15$

iv. (3,2)

$6(3) + 4(2) = 34\\18+8=34\\26\neq 34$

Putting in Equation 2:

$5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15$

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

#LearnwithBrainly

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3 years ago