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zhannawk [14.2K]
3 years ago
11

50 Points mark brainly, Which system of equations is satisfied by the solution shown in the graph?

Mathematics
1 answer:
alexira [117]3 years ago
7 0
The answer is the 4th option, 
x+y=6 and x-y= -10
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What is the answer to this
son4ous [18]
I thin the answer is D.DF

6 0
3 years ago
Helpppppp plsssssssss
bekas [8.4K]

Answer:

C

Step-by-step explanation:

The right triangle on the right side of the figure has a height of 6 (two same sides lengths) and a base of 3.

x is the hypotenuse (side opposite of 90 degree angle).

We can use the pythagorean theorem to find x. The pythagorean theorem tells us to square each leg (height and base) of the triangle and add it. It should be equal to the hypotenuse square.

For this triangle it means, we square 6 and 3 and add it. It should be equal to x squared. Then we can solve. Shown below:

6^2 + 3^2 = x^2\\36 + 9 = x^2\\45 = x^2\\x=\sqrt{45}

<em>Now we can use property of radical  \sqrt{x}\sqrt{y}  =\sqrt{x*y}  to simplify:</em>

x=\sqrt{45} \\x=\sqrt{5*9} \\x=\sqrt{5} \sqrt{9} \\x=3\sqrt{5}

Correct answer is C

6 0
3 years ago
Read 2 more answers
Estimate the rate of change of the parabola at the point x = -2.
lord [1]

Answer:

it is 2 and this is myself so haha

Step-by-step explanation:


6 0
2 years ago
Shareenah has 2 paintings. The first painting is 1 3/8
Galina-37 [17]

Answer:

3/4 meters

Step-by-step explanation:

8 0
3 years ago
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
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