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kramer
3 years ago
8

Which of the following pairs of values for x and y would justify the claim that the two triangles are congruent? x = 3, y = 11 x

= 5, y = 5 x = 7, y = 9 x = 9, y = 7
answer= A, x=3, y=11
Mathematics
1 answer:
irinina [24]3 years ago
7 0

Answer:

A

Step-by-step explanation:

just got it right

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What is the slope of the given graph?
Reptile [31]

Answer:

Looks like it's 1.

Step-by-step explanation:

Since it is a positive slope and it goes up 1 unit, the slope is 1.

7 0
3 years ago
Please help and also include an explanation so I know how you got the answer.
Nastasia [14]

Answer:

x = 15

Step-by-step explanation:

We know that 4x° + 8x° make a straight angle of 180°.

4x° + 8x° = 180°

12x° = 180°

12x = 180

x = 180/12 = 15

4 0
3 years ago
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Solve 31-X + (2 x + 1)] = x - 1.
Kobotan [32]

Answer:

x = X + 33 (I am guessing "x" and "X" are not the same. "x" is a variable and "X" is a constant.)

Step-by-step explanation:

31 - X + (2x+1) = x - 1

or, 2x+1 = x - 1 - 31 + X

or, 2x - x = -1 -31 + X -1

or, x = -2 -31 + X

or, x = X - 33

6 0
2 years ago
Solve the compound inequality. Graph your solution. 2x – 2 < –12 or 2x + 3 > 7 x < –5 or x > 5 x < –5 or x > 2
Brilliant_brown [7]

Answer:

The solution region is x < –5 and x > 2

Step-by-step explanation:

We are given the inequalities, 2x–2 < –12 or 2x+3 > 7

Upon simplifying the inequalities, we get,

A. 2x-2 < -12

2x-2 < -12 i.e. 2x < -10 i.e. x < -5

B.  2x+3 > 7

2x+3 > 7 i.e. 2x > 4 i.e. x > 2

So, we get the solution is x or x>2 and the plotted region can be seen below.

4 0
3 years ago
Determine the zeroes of the polynomial
Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
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