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2 years ago

25 POINTS PICTURE INCLUDED PLEASE ANSWER I NEED HELP What expression represents the value of v?

Mathematics

1 answer:

Ad libitum [116K]2 years ago

7 0

Answer:

that is an obtuse angle and mixed with an aacute, you answer should be 45 or 50 degrees please correct me if im wrong.

Step-by-step explanation:

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Line Equation from Two Points

Verdich [7]

Hi!

We can use point-slope form to solve this.

$y-y_{1} =m(x -x_{1})$

$y_{1}$ and $x_{1}$ will be from one of the points.

First, we have to find $m$ , the slope. We can use the slope equation to get this.

$\frac{y_{1} -y_{2} }{x_{1} -x_{2} }$

Plug in your points:

$\frac{4-0}{8-(-8)} =\frac{4-0}{8+8} =\frac{4}{16} =\frac{1}{4}$

Your slope is $\frac{1}{4}$

Now plug points and slope into point-slope equation. We will use (8, 4).

$y-4=\frac{1}{4} (x-8)$

Now, if you want to get it into y = mx + b form, you have to solve for y:

$y-4=\frac{1}{4} (x-8)$

$y-4=\frac{1}{4} x-2$

$y=\frac{1}{4} +2$

Your equation is $y=\frac{1}{4} +2$

For more information on how to get the equation of a line when given two points, see here:

brainly.com/question/986503

7 0

2 years ago

If 6x+7°and 8x-17°are vertical angles, whats x

kolezko [41]

Since the two measures are vertical angles, you want to set them equal to each other :

6x+7=8x-17
-6x -6x
——————
7=2x-17
+17 +17
———————
24= 2x
X= 12
———

The answer is 12.

4 0

3 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by h and take the limit as h approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$