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8090 [49]
4 years ago
8

I need to Find the value of x

Mathematics
1 answer:
Alenkinab [10]4 years ago
5 0
So, the sum of a triangle's angles should be 180°. the 2 angles are 60°, and if you add them up, it's 120°. 180° - 120° = 60°. the angle LMN is 60°. if a triangle has all three equal angles, that means it's a equilateral triangle. an equilateral triangle is a triangle with all equal angles, and sides. side LN is 16, so the rest should be 16. so the x is 16.
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Sooo i didn’t like April at first but now i love her
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3 years ago
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Under what conditions are we allowed to add or subtract radicals together?
Liula [17]

Answer/Step-by-step explanation:

The conditions where we are allowed to subtract or add radicals together is when the radicands, which is the number inside the radical, of both radicals are the same. And also, when they both have the same indices, that is, for example, they have the same root.

Take for example the following radical expression:

4\sqrt{2} + 3\sqrt{2}

Both radicals, 4\sqrt{2} and 3\sqrt{2}, have the same radicand, \sqrt{2} , and also the same index of square root.

Therefore, they are regarded as like terms, and can be added or subtracted together.

Thus,

4\sqrt{2} + 3\sqrt{2} = 7\sqrt{2}

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3 years ago
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Which graph has a slope of 4/5
Neko [114]
One that looks like this

6 0
3 years ago
Trevor is analyzing a circle, y2 + x2 = 100, and a linear function g(x). Will they intersect?
lakkis [162]
Given a circle described by the equation:
y^2+x^2=100
and a function g(x) given by the table
\begin{center}
\begin{tabular}{ |c |c |c }
 x & g(x) \\
 -1 & -22 \\
 0 & -20  \\
1 & -18   
\end{tabular}
\end{center}

The function g(x) describes a straight line with the equation:
\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}  \\  \\  \frac{y-(-22)}{x-(-1)} = \frac{-20-(-22)}{0-(-1)}  \\  \\ \frac{y+22}{x+1} = \frac{-20+22}{0+1} = \frac{2}{1}  \\  \\ y+22=2(x+1)=2x+2 \\  \\ y=2x-20

To check if the circle and the line intersects, we substitute the equation of the line into the equation of the circle to see if we have a real solution.
i.e.
y^2+x^2=100 \\  \\ (2x-20)^2+x^2=100 \\  \\ 4x^2-80x+400+x^2=100 \\  \\ 5x^2-80x+300=0 \\  \\ x^2-16x+60=0 \\  \\ (x-6)(x-10)=0 \\  \\ x-6=0 \ or \ x-10=0 \\  \\ x=6 \ or \ x=10

When x = 6, y = 2(6) - 20 = 12 - 20 = -8 and when x = 10, y = 2(10) - 20 = 20 - 20 = 0

Therefore, the circle and the line intersect at the points (6, -8) and (10, 0).
8 0
4 years ago
Read 2 more answers
Help(Worth 20 Points)
Alik [6]

Answer:

24

Step-by-step explanation:

1.-12

2.-12×2=-24

3.24

4 0
3 years ago
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