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Blababa [14]
4 years ago
13

Solve for a in the diagram below.

Mathematics
2 answers:
Marat540 [252]4 years ago
8 0

Answer:

20

Step-by-step explanation:

Angles form a right angle, hence they add up to 90

x + 2x + x + 10 = 90

4x + 10 = 90

4x = 80

x = 20

Tresset [83]4 years ago
7 0

Answer:

20

Step-by-step explanation:

We can tell from the little square in the corner that this is a 90 degree angle, so all three of the expressions of x should add up to 90:

x + 2x + (x + 10) = 90

Combine like terms:

4x + 10 = 90

4x = 80

Divide by 4:

x = 80/4 = 20

The answer is 20.

Hope this helps!

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3x+5y = 15 find the slope of a line perpendicular to each given line
gladu [14]

Answer:

Slope = 5/3

Step-by-step explanation:

See attached image

5 0
2 years ago
The discriminant for the equation f(x)=6x^(2)-x+c is 73. What is the value of c?
Andrews [41]

Answer:

c = -3

Step-by-step explanation:

The discriminant  is 73, that means we have:

(-1)^2-4*6*c=73

then: 1-24c=73

or -24c= 73 -1

-24c = 72

Then c = 72/-24= -3

The answer is -3

Hope that useful for you.

5 0
3 years ago
Evaluate the function g(x) = –2x2 + 3x – 5 for the input values –2, 0, and 3.
Genrish500 [490]

Answer:

 g(-2) = -15, g(0) = -9, g(3) = -18

Step-by-step explanation:

g(-2) = -2*2 + 3(-2) - 5 g(-2) = -4 - 6 - 5g(-2) = -15g(0) = -4 - 0 - 5g(0) = -9g(3) = -4 - 3(3) - 5g(3) = -4 - 9 - 5g(3) = -18

I HOPE THIS HELPS

4 0
2 years ago
Find the illegal values of c in the multiplication statement c^2-3c-10/c^2+5c-14 times c^2-c-2/c^2-2c-15
Citrus2011 [14]

Answer: c ≠ {-7, -3, 2, 5}

<u>Step-by-step explanation:</u>

  \frac{c^{2} - 3c - 10}{c^{2} + 5c - 14} * \frac{c^{2} - c - 2}{c^{2} - 2c - 15}

= \frac{(c - 5)(c + 2)}{(c + 7)(c - 2)} * \frac{(c - 2)(c + 1)}{(c - 5)(c +3)}

restrictions:

c + 7 ≠ 0           c - 2 ≠ 0          c - 5 ≠ 0          c + 3 ≠ 0

    c ≠ -7                 c ≠ 2               c ≠ 5                c ≠ -3

3 0
4 years ago
Brainliest if correct!<br>i need help w this. need it by wednesday
skelet666 [1.2K]

Answer:

I need help to this thing even given me what I need

5 0
2 years ago
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