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Sophie [7]
2 years ago
11

(-30) x20 = please tell ​

Mathematics
1 answer:
Afina-wow [57]2 years ago
4 0

Answer:

-600

Step-by-step explanation:

-3 x 2 is -6

then add the zeros

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Determine the value of w and x in the figure below. (Thanks for the help I really appreciate it!)​
storchak [24]
Triangle HAM looks like an isosceles triangle, and we know that in an isosceles triangle 2 of the angles are the same.

so :
180 - 106 = 74 ( sum of the other 2 angles )
74/2 = 37 ( sum of angle HMA / HAM )

angles on a straight line add up to 180.
180-37=143 ( angle x )

since triangle YMH is also an isosceles,
180-37=143
143/2=71.5 ( angle w )
7 0
3 years ago
When will the ball hit the ground
Likurg_2 [28]

Sorry but you didnt add a picture nor description. If you can, please re-ask this question with the following:

Details to your question

A picture or drawing

Thanks - Madilyn.

5 0
3 years ago
Read 2 more answers
(5 x 10^2)^−2<br> i dont get it
ss7ja [257]

Answer:-250,000

Step-by-step explanation:

3 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
2 years ago
What number should be added to both sides of the equation to complete the square? x^2 – 6x = 5
NeTakaya
You will add ( \frac{-6}{2}  )^{2}
(-3)^{2}
9
5 0
3 years ago
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