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3 years ago

Find the value of x.

Mathematics

2 answers:

jek_recluse [69]3 years ago

5 0

Answer:

41 degrees

Step-by-step explanation:

180-90-49 = 41

IrinaK [193]3 years ago

3 0

Answer:

the answer is 49

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Square root of 32 x square root of 1 over 18

lisov135 [29]

<h3>Solution:</h3>

$\sqrt{32} \times \sqrt{ \frac{1}{18} } \\ = \sqrt{32} \times \frac{ \sqrt{1} }{ \sqrt{18} } \\ = \sqrt{32} \times \frac{1}{ \sqrt{18} } \\ = \frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}{ \sqrt{2 \times 3 \times 3} } \\ = \frac{ \sqrt{ {2}^{2} \times {2}^{2} \times 2} }{ \sqrt{2 \times {3}^{2} } } \\ = \frac{2 \times 2 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4}{3}$

<h3>Answer:</h3>

$\frac{4}{3}$

<h3>Hope it helps...</h3>

ray4918 here to help

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3 years ago

The X- and y-coordinates of point P are each to be chosen at random from the set of integers 1 through 10.

bagirrra123 [75]

Answer:

Ok, as i understand it:

for a point P = (x, y)

The values of x and y can be randomly chosen from the set {1, 2, ..., 10}

We want to find the probability that the point P lies on the second quadrant:

First, what type of points are located in the second quadrant?

We should have a value negative for x, and positive for y.

But in our set; {1, 2, ..., 10}, we have only positive values.

So x can not be negative, this means that the point can never be on the second quadrant.

So the probability is 0.

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3 years ago

if mrs wright had 90 pieces of candy how many pieces would she need in order to give each student the same amount as the 27 stud

Alisiya [41]

Answer:

90÷27= 3. hope it helps

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