What is it fast tell me

Answers

Answer 1

Answer: d is the answer to this

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What is the value of root 12+root 12+root12....
...............to infinite

Answers

x=root(12+root(12+root(...)))
x^2=12+root(12+root(12+root(...)))
x^2=12+x
x^2-x-12=0
(x+3)(x-4)=0
X=4

A circle with radius of 1 cm sits inside a 11 cm x 12 cm rectangle.What is the area of the shaded region?
Round your final answer to the nearest hundredth.

Answers

Answer:

128.86 square centimetres

Step-by-step explanation:

I will assume that the shaded region is the area in between the circle and the rectangle.

All we have to do is find the area of the circle and subtract it from the area of the rectangle.

Area of a circle is given as:

$A = \pi r^2$

where r = radius

The radius of the circle is 1 cm, so, the area of the circle is therefore:

$A = \pi * 1^2$ = 3.14 square centimetres

The area of a rectangle is given as:

A = L * B

The dimensions of the rectangle are 11 cm by 12 cm. The area of the rectangle is:

A = 11 * 12 = 132 square centimetres

The area of the shaded part is therefore:

A = 132 - 3.14 = 128.86 square centimetres

a city has two water towers. one tower holds 8.4 x 103 gallons of water and the other tower holds 9.5 x 104 gallons of water. what is the combined water capacity of the two towers in scientific notation?

Answers

Total capacity the tower holds 1.034 x 10^5 gallons of water

One tower holds 8.4 x 10^3 gallons of water

Other tower holds 9.5 x 10^4 gallons of water

The combined water capacity of the two towers in scientific notation is,

Total capacity the tower holds = 9.5 x 10^4 + 8.4 x 10^3

= 9.5 x 10^4 + 0.84 x 10^4

= 10.34 x 10^4

= 1.034 x 10^5

To learn more about Capacity click here:

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Which undefined terms are needed to define a line segment? A. Plane, point

B. Line, endpoint

C. Point, line

D. Segment, point

Answers

The answer is c point line

Thats the wrong answer

Find the real or imaginary solutions of the equation by factoring. x^4-8x^2=-16

Answers

ANSWER
$x=2$ or $x=-2$
EXPLANATION

The given equation is
${x}^(4) - 8 {x}^(2) = - 16.$

We set everything equal to zero to obtain,

${x}^(4) - 8 {x}^(2) + 16 = 0$

We rewrite the leading term to obtain,

${( {x}^(2) )}^(2) - 8( {x}^(2)) + 16 = 0$

This has now become a quadratic equation in
${x}^(2)$

We split the middle term to get,

${( {x}^(2) )}^(2) - 4 {x}^(2)- 4x^2+ 16 = 0$

We factor to get,

${x}^(2) (x^2-4)-4( {x}^(2)-4) = 0$

We factor further to get,

$( {x}^(2) -4)(x^2-4)= 0$

This gives

$( x-2)(x+2)(x-2)(x+2)= 0$

$( x-2)^2(x+2)^2= 0$

$x=2$ or $x=-2$

60 POINTS !!!!!! PLEASE HELP