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mamaluj [8]
3 years ago
15

HELLLLPPPPPPPP☝️☝️☝️☝️☝️!!!!

Mathematics
1 answer:
8090 [49]3 years ago
5 0
Money collected independent
tickets sold dependent
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One side if the labtop is 9.5 inches and the other side is 14 inches. Using the Pythagorean Theron, what is the length of the di
Digiron [165]

(side 1)² + (side 2)² = (diagonal)²

(9.5² + (14)² = (diagonal)²

90.25 + 196 = (diagonal)²

286.25 = (diagonal)²

√286.25 = √(diagonal)²

17 = diagonal

Answer: 17 inches

3 0
3 years ago
Use the distributive property to remove the parentheses.<br> (v-2)9
Sunny_sXe [5.5K]

Answer:

9v-18

Step-by-step explanation:

.........................................

5 0
3 years ago
How do humans interact with the geosphere? Explain
tresset_1 [31]

Answer:

there are many ways

Step-by-step explanation: we walk on the crust of our earth interact with rocks, and some glasses are made of molten lava. which is also from a specific layer of earth (sorry, dont remember which) and some people drill in the mines and maybe even get lower than the crust. i hope this helps :/

8 0
2 years ago
Write the equation of the line that passes through the points (4, -6) and (-2,3).
Butoxors [25]
In point slope form, the answer would be y = - 3/2 x it’s one fraction
5 0
3 years ago
Read 2 more answers
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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