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2 years ago

4.72x10^10 Please I need help fast

Mathematics

1 answer:

vodka [1.7K]2 years ago

5 0

Answer:

The answer is 47200000000

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Plz help due tonight

MariettaO [177]

Sum of the interior angles of any triangle equal to 180 degrees thus :

60 + 70 + 8x + 2 = 180

8x + 132 = 180

Subtract both sides 132

8x + 132 - 132 = 180 - 132

8x = 48

Divide both sides by 8

8x ÷ 8 = 48 ÷ 8

6 0

3 years ago

Read 2 more answers

A small square of side 2x is cut from the corner of a rectangle with a width of 10 centimeters and length of 20 centimeters. Wri

love history [14]

Find the area of the rectangle. You can find the area of a rectangle with the following formula:

$A = wl$

w = width, and l = length.

Plug in your width and length:

$20 * 10 = 200$

The area of the rectangle is 200 square centimeters.

The area of a square is also given by width * length. Multiply the sides together:

$2x * 2x = 4x^{2}$

The area of a square is 4x^2.

Because you're cutting away the square from the rectangle, you will subtract the area of the square from the area of the rectangle. The following equation will be your answer:

$200 - 4x^{2} cm^{2}$

6 0

3 years ago

N is an integer.<br>Write the values of n such that -15 < 3 < 6

Thepotemich [5.8K]

Answer:

Step-by-step explanation:

n could be : -14, -13, -12... -1, 0 , 1, 2, 3 ,4,5

3 0

3 years ago

Find the area of a tank which measures 25cm and 10cm

laiz [17]

Answer:

Area = 250

Step-by-step explanation:

25 x 10

Length * Width

5 0

2 years ago

Create a matrix that is equal to F+G. The first matrix below is named F and the second matrix below is named G. Name the new mat

likoan [24]

F+G:

$F+G=\begin{bmatrix}{-1.8} & {-8.6} & {} \\ {2.85} & {-1.4} & {} \\ {-1.8} & {5.1} & {}\end{bmatrix}+\begin{bmatrix}{1.32} & {-1.9} & {} \\ {2.25} & {0.0} & {} \\ {-6.2} & {1.4} & {}\end{bmatrix}$

Then, add the elements that occupy the same position:

$H=\begin{bmatrix}{-1.8+1.32} & {-8.6+(-1.9)} & {} \\ {2.85+2.25} & {-1.4+0.0} & {} \\ {-1.8+(-6.2)} & {5.1+1.4} & {}\end{bmatrix}$

Solve

$H=\begin{bmatrix}{-0.48} & {-10.5} & {} \\ {5.1} & {-1.4} & {} \\ {-8} & {6.5} & {}\end{bmatrix}$

So, we find the element at address h31:

$H=\begin{bmatrix}{h11} & {h12} & {} \\ {h21} & {h22} & {} \\ {h31} & {h32} & {}\end{bmatrix}$

In this case, position h31 is - 8.0

8 0

1 year ago