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Aleks04 [339]
2 years ago
9

When dividing a decimal by 10³, how is the decimal point moved?

Mathematics
1 answer:
shtirl [24]2 years ago
8 0

Answer:

3 digits to the left side

Step-by-step explanation:

The decimal point will move 3 digits to the left side when dividing a decimal by 10³.

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Exterior Angles of A Triangle
gladu [14]
The exterior angle is supplementary to the adjacent angle (they form a linear pair) and is the sum of the two angles in the triangle that are not adjacent to it. For example, in number 2, the exterior angle is 28+40=68 degrees. You can use the other method to check. 68 degrees is supposed to be in a linear pair with 112 degrees, which it is. So the answer to number 2 would be 68 degrees.

Try doing this for the rest of them
7 0
2 years ago
What is the equation of the line that passes between (-1, -7) and has a slope of 2
cricket20 [7]

Answer:

y = 2x - 5

Step-by-step explanation:

3 0
3 years ago
HELP WILL GIVE BRANLIEST! WILL RATE! AND SAY THANKS!!!
Drupady [299]

Answer:

9.57 inch

Step-by-step explanation:

(4/3)×pi×8³ = 14×16×h

h = (4/3)×pi×8³ ÷ (14×16)

h = 64pi/21 inch

Or, 9.57 inch (3 sf)

5 0
3 years ago
What number has two more tens then 11
777dan777 [17]
The answer is 31. 10+10+20+1=31
5 0
3 years ago
How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%
blondinia [14]

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2

Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
3 years ago
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