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3 years ago

What is the answer to this problem!?

Mathematics

1 answer:

stepladder [879]3 years ago

4 0

I cant read it...

ill answer if know

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What is the equatipn of the linear graph? (-4, 6). (1, -9)

dusya [7]

Answer:

y = -3x - 6

Step-by-step explanation:

first, find the slope

(-4, 6)

(1, -9)

remember the equation for the slope

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

insert given points

$x_{1} = -4\\y_{1} = 6\\\\x_{2} = 1\\y_{2} = -9$

((-9) - (6)) / ((1) - (-4))

simplify

( -9 - 6) / (1 + 4)

= -15/ 5

= -3

now insert it into the equation of the line

remember, the equation of the line is

y = mx + b

where "m" is the slope

and "b" is the y-intercept

to find "b", you just have to plug in a point that is found on the line

we know that (-4, 6) and (1. -9) is found on the line, so let's just use (-4, 6) for this problem

y = mx + b

substitute in the slope of this line

y = -3x + b

now substitute in the given points

6 = -3 * -4 + b

and solve it like any algebra problem, by inverse operations and simplifying

6 = 12 + b

-12 -12

-6 = b

so the equation of the line is

y = -3x -6

3 0

3 years ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

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

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

3 years ago

In the diagram below of parallelogram ROCK, m angle is 70 degrees and m angle ROS is 65 what is the value of x

il63 [147K]

Answer:

135

Step-by-step explanation:

5 0

3 years ago

What is the Graph y=x+4

Marizza181 [45]

Answer:

It is a line with a positive slope and positive intercept

Step-by-step explanation:

If you want it in SF, -x + y = 4

3 0

3 years ago

Simplify. 54+4(34−12)2 Enter your answer in the box.

gtnhenbr [62]

Answer:

230

Step-by-step explanation:

54+4×22×2

54+88×2

54+176

230

3 0

3 years ago

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