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Eduardwww [97]
3 years ago
11

Please Help! Solve this system of equations { 1/2x+6y+2 y=x+15

Mathematics
1 answer:
REY [17]3 years ago
4 0

I dont know of you wanted the graph too so here

everything is in the pictures here, also sorry if I am wrong

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What are all prime numbers x which:<br> x&lt;10
ivanzaharov [21]

Answer:

5,7,3,2

Step-by-step explanation:

It's pretty simple: a prime number is a number that can *only* be divided by the number 1 and it's own.

I just checked all the numbers from 1 to 10 in order to solve this problem.

If you have any questions about the way I solved it, don't hesitate to ask!

8 0
3 years ago
The vertex angle of an isosceles triangle is 20° less than the sum of the base angles. Which system of equations can be used to
WINSTONCH [101]

The answer to the above question can be explained as under -

We know that, the sum of angles of triangle is 180°.

So, vertex angle plus base angles are equal are equal to 180°.

Let the vertex angle be represented by "v" and base angles be represented by "b".

Thus,  v + b + b = 180°

So,  v + 2b = 180°

Next, the question says, the vertex angle is 20° less than the sum of base angles.

Thus, 2b - 20° = v

<u>Thus, we can conclude that the correct option is A) v + 2b = 180°, 2b - 20° = v</u>

6 0
3 years ago
¾ is considered a real and rational number. true or false?
allochka39001 [22]
IT is True. 
<span>¾ is considered a  real number</span>
3 0
3 years ago
Read 2 more answers
The length of a rectangle is 3 inches longer then twice its width, where w is the width of the rectangle the area of the rectang
vaieri [72.5K]
A = w(2w + 3)
90 = 2w^2 + 3w
2w^2 +3w - 90 = 0
(w-6)(2w+15) = 0        (TRINOMIAL FACTORING)
w = 6 inch                   ( it can't be -15/2 because lengths can't be negative)
l = 2w + 3
  = 15 inch
6 0
3 years ago
5<br> 0 = - in<br> =<br> in quadrant II<br> Given cos =<br> Find sin<br> 3
____ [38]

Answer:

sinΘ = \frac{2}{3}

Step-by-step explanation:

using the identity

sin²x + cos²x = 1  ( subtract cos²x from both sides )

sin²x = 1 - cos²x ( take square root of both sides )

sinx = ± \sqrt{1-cos^2x}

given

cosΘ = - \frac{\sqrt{5} }{3} , then

sinΘ = ± \sqrt{1-(-\frac{\sqrt{5} }{3})^2 }

        = ± \sqrt{1-\frac{5}{9} }

        = ± \sqrt{\frac{4}{9} }

        = ± \frac{2}{3}

since Θ is in quadrant II where sinΘ > 0 , then

sinΘ = \frac{2}{3}

7 0
2 years ago
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