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3 years ago

A circle has a diameter of 42 mm. What is the circumference of this circle? (use 22/7 for pi; show your work in numbers)

Mathematics

1 answer:

stellarik [79]3 years ago

8 0

Answer:

132

Step-by-step explanation:

Find Radius (r)

42/2 = 21

Plug 21 into Circumference Equation

2πr (In this case, π is 22/7)

Simplify

42 (22/7)

132

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What Is the Slope And Y Intercept Of This Line? Y = X -5

asambeis [7]

Answer:

slope = 1 , y- intercept = - 5

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

y = x - 5 is in this form

with slope m = 1 and y- intercept c = - 5

6 0

3 years ago

How many 1/4 inch cubes does it take to fill a box with width 3 1/4 inches, length 4 1/2 inches, and height 2 3/4 inches?

Anastaziya [24]

1. 15 2. 18. 3. 11 is right

4 0

3 years ago

Read 2 more answers

Answer pls The hint part also

tatiyna

4x+60+ x+15=180
5x+75=180
5x=105
x=21
Two linear angles are supplementary thereom

3 0

3 years ago

PLS AWNSER ASAP!!Which expression can be used to determine the length of segment ZY? On a coordinate plane, triangle X Y Z has p

MArishka [77]

Sqrt(8^2 + 3^2)

Y is 3,4
Z is -5,1

3-(-5) = 8 (difference between x coords)
4-1 = 3 (difference between y coords)
Equation is: sqrt( (xdiff)^2 + (ydiff)^2 )

7 0

3 years ago

Find a cubic function with the given zeros.

Fed [463]

Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

$f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 - 2x^2 - 2x - 4$ .

$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

8 0

3 years ago