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SVETLANKA909090 [29]

2 years ago

12

X^2-16x+64=-8 take the square root of both sides

1 answer:

dimulka [17.4K]2 years ago

3 0

I think the answer is x2 +64 but I hope you get it right good luck ! I hope it is 64 thanks

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In matrix theory, many of the familiar properties of the real number system are not valid. If a and b are real numbers, then ab

kozerog [31]

Answer:

B= $\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right]$

Step-by-step explanation:

Let's do the multiplication AB.

If A= $\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right]$

then the first row of A is= (1 0) by the first column of B= (0 0) is equal to zero.

the first row of A is= (1 0) by the second column of B= (0 1) is equal to zero too because 1.0+0.1=0.

the second row of A is= (0 0) by any colum of B is equal to zero too.

So we have found an example that works!

7 0

3 years ago

Help! Picture attached

schepotkina [342]

Answer:

minimum maximum parabolas are equal then because of the circumference

Step-by-step explanation:

5 0

2 years ago

What characteristic of a parabola can be easily identified when the factored form of a quadratic is given?

Shkiper50 [21]

D) the x intercept as you let each bracket equal to 0 to find the x coordinate

3 0

2 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

2 years ago

Elaine Mellon earns $17.80 per hour as an electric billing specialist for a legal clinic. How many hours per week must she work

Nuetrik [128]

Answer:

17.80/676.4= $38

Step-by-step explanation:

8 0

3 years ago