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3 years ago

I need help with this equation 20 points help quick

Mathematics

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

The equation of the graph after translating one unit to the left is;

$y = \left | \dfrac{x}{2} - \dfrac{3}{2} \right |+3$

Step-by-step explanation:

The given equation is $y = \left | \dfrac{1}{2}\cdot x - 2 \right |+3$

We note that minimum value of y = 3, where 1/2·x - 2 = 0 and x = 4

Therefore, in moving one unit to the left, we have at the y-intercept where slope of the graph has become inverted (reflection of the real graph) we add one to the x value as follows;

$y = \left | \dfrac{1}{2}\cdot (x+1) - 2 \right |+3 = \left | \dfrac{x}{2} + \dfrac{1}{2} - 2 \right |+3 = \left | \dfrac{x}{2} - \dfrac{3}{2} \right |+3$

The equation of the graph becomes;

$y = \left | \dfrac{x}{2} - \dfrac{3}{2} \right |+3$ .

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General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

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3 years ago

What are the roots of this quadratic equation?

pickupchik [31]

Answer:

Step-by-step explanation:

im the best in the world

5 0

1 year ago

An observer in a hot air balloon sights a building that is 50 m from the balloon's launch point. The balloon has risen 165 m. Wh

Kaylis [27]

Notice the picture,
recall your SOH, CAH, TOA
$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}$

you have,
opposite side, 165
adjacent side, 50
and the angle

that means, we'll need Mrs. tangent
thus
$\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50}
\\ \quad \\
 tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right]
\\ \quad \\
\theta=tan^{-1}\left[ \cfrac{165}{50}\right]
\\ \uparrow \\
 \textit{angle of elevation}\iff\textit{angle of depression}$

4 0

3 years ago

Evaluate each expression.

Kitty [74]

The fraction form is 125/343

the decimal form is 0.36

8 0

2 years ago

Read 2 more answers

What is y=-4+15 and y=8x-33 in substitution

Hoochie [10]

8x - 33 = - 4x + 15
12x = 48
x = 4

Then substitute x back into a original equation to find y

y = (8x4) - 33
y = 32 - 33
y = - 1

So the answer is x=4 and y=-1

3 0

3 years ago