How many feet are in 120 inches? Enter your answer in the box. ft

sid walked 6 miles toward his home which is 3/5 of the total distance. What is the total distance to his home?

tensa zangetsu [6.8K]

Sid walked 6 miles, which is 3/5 of the total distance.

Divide 6 with 3

6/3 = 2

2 miles is the walking distance for 1/5

Multiply 5 to both sides

2 (x 5) = 1/5 (x 5)

10 = 1

The total distance is 10 miles

hope this helps

4 0

3 years ago

•Use the Pythagorean Theorem c^6 = a2+ b2 •Show you work to find each missing side •

zhannawk [14.2K]

Answer:

hypotenuse = 10

Step-by-step explanation:

If you can tell at first glance it is a 3,4,5 triangle (6,8,10)/2=(3,4,5). Then you know that c has to equal 5.

If not,

a^2+b^2=c^2

6^2+8^2=c^2

36+64=c^2

100=c^2

10=c

6 0

3 years ago

Jan received -22 points on her exam. She got 11 questions wrong out of 50 questions. How much was Jan penalized for each wrong a

svp [43]

2 points for each incorrect answer.

6 0

3 years ago

Read 2 more answers

What is the value of x? Enter your answer in the box.

Tamiku [17]

It's a hexagon.

The formula of the sum of internal angles in the polygon:

$S=(n-2)\cdot180^o$

Therefore

$112^o+133^o+128^o+100^o+120^o+x^o=(6-2)\cdot180^o\\\\593^o+x^o=720^o\ \ \ \ |-593^o\\\\x^o=127^o$

3 0

3 years ago

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