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irakobra [83]
3 years ago
11

PLEASE I NEED HELP WITH MATH HW WITH EXPLANATION!!! Question: Find the value of the varible

Mathematics
1 answer:
Marizza181 [45]3 years ago
5 0
<h3>Answer:  w = 31</h3>

===================================================

Explanation:

For any triangle, the interior angles always add to 180

(angle1)+(angle2)+(angle3) = 180

(w) + (4w-6) + (w) = 180

(w+4w+w) - 6 = 180

6w - 6 = 180

6w - 6+6 = 180+6 .... adding 6 to both sides

6w = 186

6w/6 = 186/6 ..... divide both sides by 6

w = 31

------------------------------------

Extra info:

This value of w leads to

4w-6 = 4*31-6 = 118

The three angles of the triangle are: 31, 118, 31

Because we have two congruent angles, this means the triangle is isosceles. The sides opposite the congruent angles are the same length.

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Write an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the line described by y=1/2
Marat540 [252]

Given:

The point, (4, -3)

The line,

y=\frac{1}{2}x+5

To find an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the given line:

The slope of the line is,

m=\frac{1}{2}

Since the given line is parallel to the new line, so the slope will be same for the both.

Using the point-slope formula,

y-y_1=m(x-x_1)

Substitute the point and slope we get,

\begin{gathered} y-(-3)=\frac{1}{2}(x-4) \\ y+3=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-2-3 \\ y=\frac{1}{2}x-5 \end{gathered}

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y=\frac{1}{2}x-5

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(a) Find the slope of the curve y= x^2 - 2x - 3 at the point ​P(2​,​ -3) by finding the limit of the secant slopes through point
gayaneshka [121]

Answer:

Step-by-step explanation:

P=(2,-3)

Q=(a,a²-2a-3)

\Delta y=a^2-2a-3-(-3)=a^2-2a\\\Delta x=a-2\\\\\displaystyle \lim_{a \to 2} \dfrac{a^2-2a-3-(-3)}{a-2}  \\\\=\lim_{a \to 2} \dfrac{a^2-2a}{a-2}  \\\\=\lim_{a \to 2} \dfrac{a(a-2)}{a-2}  \\\\=\lim_{a \to 2} \dfrac{a}{1}  \\\\=\lim_{a \to 2} a  \\\\=2\\

Proof:

y'=(x²-2x-3)'=2x-2

y'(2)=2*2-2=2

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