What are the real or imaginary solutions of the polynomial equation? x^3-8=0

5. If 1/4 inch represents 5 yards, the number of inches needed to show a football field 100yards long is:

alexdok [17]

Answer:

The answer is 20/80.

Step-by-step explanation:

1/4=5 yards

? =100 yards

100÷5=20

1/4×20

=20/80.

PLEASE MARK ME AS THE BRAINIEST.

4 0

2 years ago

The formula for the future value V (in dollars) of an investment earning simple interest is V=p+prt, where p (in dollars) is the

zhannawk [14.2K]

a)

V=p+prt

now we solve for P

V=P (1+rt)

Divide both sides by (1+rt)

P=V÷(1+rt)...answer

b)

P=V÷(1+rt)

P=3,000÷(1+0.06×5)

P=2,307.69

5 0

3 years ago

Can someone pls help me with these I’ve asked this question 3 times already :( will mark brainiest

Tresset [83]

1: cell

2: distinction

3: assumption

4: foliage

5: commision

6: viewpoint

7: considerable

8: membrane

9: cell

10: final

11: viewpoint

12: considerable

uwu

Edit: added 9-12 didn't see them until I accidentally scrolled down, lol.

7 0

3 years ago

Read 2 more answers

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

6 0

3 years ago

There are 20 cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all t

blagie [28]

Answer:

The answer is C/85,000

Step-by-step explanation:

6 0

2 years ago