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saul85 [17]
3 years ago
8

Can someone please help me With explanation please ty

Mathematics
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

Basically asking you which two numbers multiplied together is 20. Keep in mind this is a triangle so you'll need to divide the product by 2. In this case, it would be C.4 and 10

4 time 10 is 40

40/2 = 20

Step-by-step explanation:

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Answer:

When the inputs are 1 and 0, the output is zero.

7 0
3 years ago
Factor as the product of two binomials x^2-3x-10
cestrela7 [59]

Answer:

(x-5)(x+2)

Step-by-step explanation:

Hello there!

Your expression there has a highest power of x^{2}, so there will be two constants that you need to find.

Your last term is -10, so the constants will have a product of -10. Also your middle term is -3x, so the terms will add up to -3

-5 and 2 fit the mold.

Have a great day!

If I am most helpful, mark me brainliest!

3 0
3 years ago
Evaluate. <br> 6[5(3-9)-1]+2/7(8-2)+4
GREYUIT [131]
If you would like to evaluate 6 * [5 * (3 - 9) - 1] + 2 / 7 * (8 - 2) + 4, you can calculate this using the following steps:

6 * [5 * (3 - 9) - 1] + 2 / 7 * (8 - 2) + 4 = 6 * [5 * (-6) - 1] + 2/7 * 6 +  4 = 6 * [- 30 - 1] + 12/7 + 4 = 6 * [- 31] + 12/7 + 4 = - 186 + 12/7 + 4 = - 182 + 12/7 = - 1274/7 + 12/7 = - 1262/7

The correct result would be - 1262/7.
6 0
4 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
600000 80000 10 expanded. form
kirza4 [7]
600000+80000+10=680,010 six hundred eighty thousand ten
7 0
3 years ago
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