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4 years ago

6

A cylinder has a radius of 6 inches and a height of 2 feet. Which of the following should be done first in order to find the vol

ume of the cylinder?
convert the radius and the height to the same unit
determine the diameter of the circle
calculate the circumference of the circle
multiply the circumference of the circle by the height of the cylinder

1 answer:

zvonat [6]4 years ago

7 0

The significant components, the radius and the height, are in inches and feet respectively, you'd need to convert them both to the same unit, whichever it may be, meters or feet or inches or millimeters, so long is the same, then use the volume of a cylinder equation to get it.

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Does the system shown always, sometimes, or never have a solution when a >b?

Rama09 [41]

Answer:

sometimes

Step-by-step explanation:when a >b?

ax+y=8

bx + y = 5

always

sometimes

O

never

5 0

3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

<u>case B)</u> $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

<u>case C)</u> $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

<u>case D)</u> $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

<u>the answer is</u>

$f(x)=(x-2)(x+2)$

6 0

3 years ago

Read 2 more answers

vaieri [72.5K]

1 : 100,000
Because there are 100cm in a m and 1000m in a km
100 x 1000 = 100,000

3 0

3 years ago

Pls help this is very easy

anastassius [24]

Answer:

20/9

Step-by-step explanation:

Alright, so by using the "keep me, change me, turn me over" method - we can easily solve this:

$\frac{8}{9}\div \frac{2}{5}\\=\frac{8}{9}\times \frac{5}{2}\\=\frac{4}{9}\times \frac{5}{1}\\=\frac{4\times \:5}{9\times \:1}\\=\frac{20}{9}$

<em>I hope I was of assistance!</em> <em><u>#SpreadTheLove <3</u></em>

4 0

3 years ago

Write an equation of the line that has a slope of 2 and y-intercept 1 in slope-intercept form.

Vlada [557]

Y = mx + b

in this case m = 2 and b = 1 so
equation
y =2x + 1

hope this helps

4 0

3 years ago

Read 2 more answers