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3 years ago

8

Which graph represents the function f(x) = x-2?

2 answers:

densk [106]3 years ago

5 0

Answer:

click in photo

nejdjjd

nxndjdbbdjf

MatroZZZ [7]3 years ago

3 0

lấy x=0 ta có y= -2 => A(0;-2)

lấy y=0 ta có x=2 => B(2;0)

nối 2 điểm A và B ta có đồ thị:

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The graph below has the same shape. What is the equation of the blue graph????????

vesna_86 [32]

Answer: The answer is a ahahahha thats funny im in 8th grade and i figured that out and we are not even doing that yet its just infering

4 0

3 years ago

Simplify the expression: 2[(8−4)5÷8]

nignag [31]

Answer:

5.

Step-by-step explanation:

2[(8−4)5/8]

= 2[4*5 / 8]

= 2 * 2.5

= 5.

8 0

3 years ago

Mr.Gorman spent $65 on the following art supplies

Bess [88]

Part A: $2.50 * 5 pads = $12.50

$5.75 * 6 pkg markers = $34.50

$12.50 + $34.50 = $47.00 spent on pads and markers

Total spent $65.00 - $47.00 spent on pads and markers = $18.00 left for jars of paint

$18.00 divided by $4.50 = 4 jars of paint that he bought

PART B. The answer is 12 colored pencils.

$18 total spent 3 paint brushes = $9.00

$18 - $9= $9.00 left for color pencils

$9.00 divided by $0.75= 12 colored pencils

the answer is 12 colored pencils

5 0

3 years ago

How to calculate roots when the equation is non quadratic?

likoan [24]

Solve like a linear algebra equation
$y = mx + b$
use for slope
$m = y1 - y2 \div x1 - x2$
the b is intersept

5 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

4 years ago