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vodomira [7]
3 years ago
11

Please someone help me with this I am very confused

Mathematics
1 answer:
Elis [28]3 years ago
4 0
8 \sqrt{3} *4 \sqrt{15} = 8*4 \sqrt{3*15}  \\  \\ 32 \sqrt{45} = 32 \sqrt{9*5} =32*3 \sqrt{5}  \\  \\ = 96 \sqrt{5} 

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Solve for all values of x by factoring
oksian1 [2.3K]

Answer:

x= 0,5

Step-by-step explanation:

You have to move all the terms to the left side and set them equal to zero. Then set each factor equal to zero.

Hope this helped you!

⭐ If it helped you please give brainliest! ⭐

7 0
2 years ago
What is the equation of the line of best fit for the following data? Round the slope and y-intercept of the line to three decima
Nadusha1986 [10]

Answer:

d

Step-by-step explanation:

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8 0
2 years ago
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What is parallel to 2x-y=1 and contains (4, -6)? Please show your work!
Contact [7]

k:\ y=m_1x+b_1\\\\l:\ y=m_2x+b_2\\\\l\ ||\ k\iff m_1=m_2\\\\\text{We have}\\\\k:\ 2x-y=1\ \ \ |-2x\\-y=-2x+1\ \ \ \ |\cdot(-1)\\y=2x-1\to m_1=2\\\\l:\ y=m_2x+b\\\\l\ ||\ k\iff m_2=2\\\\l:\ y=2x+b\\\\\text{The line l contains (4, -6)}\\\\\text{Put the values of coordinates to the equation of line l}\\\\-6=2(4)+b\\-6=8+b\ \ \ \ \ |-8\\b=-14\\\\Answer:\ y=2x-14\to 2x-y=14

5 0
3 years ago
a jar contains 6 red jelly beans, 6 green beans, and 6 blue jelly beans if we choose a jelly bean, then another jelly bean witho
seropon [69]
<h3>Answer: 5/51</h3>

======================================================

Explanation:

We have 6 green out of 6+6+6 = 18 total

The probability of getting green is 6/18 = 1/3.

After selecting that green jelly bean and not putting it back, we have 6-1 = 5 green out of 18-1 = 17 total.

The probability of selecting another green is 5/17.

Multiply the two fractions 1/3 and 5/17

(1/3)*(5/17) = (1*5)/(3*17) = 5/51

The probability of selecting two greens in a row is 5/51 where we do not put the first selection back. We also do not replace the green jelly bean with some other identical copy.

Note: 5/51 = 0.098039 approximately

8 0
3 years ago
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lesya692 [45]

Answer:

{ \tt{g(x) = 5x - 4}}

• solve when x is a - 1:

{ \tt{g(a - 1) = 5(a - 1) - 4}} \\  \\ { \tt{g(a - 1) = 5a - 5 - 4}} \\  \\ { \boxed{ \tt{g(a - 1) = 5a - 9 \: }}}

8 0
2 years ago
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