Ask question

Ask question

All categories

White raven [17]

2 years ago

11

Find the slope and y-intercept. Also, determine if it is Proportional or Non Proportional

1 answer:

lubasha [3.4K]2 years ago

8 0

Answer:

Step-by-step explanation:

Slope of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

m = $\frac{y_2-y_1}{x_2-x_1}$

From the table attached,

Two points lying on the graph are (1, 12) and (2, 24).

Slope of the line will be,

m = $\frac{24-12}{2-1}$

Slope = 12

Since equation of a line passing through a point (x', y') and slope 'm' is,

y - y' = m(x - x')

Therefore, equation of the line passing through (1, 12) and slope 12 will be,

y - 12 = 12(x - 1)

y = 12x - 12 + 12

y = 12x

By comparing this equation with the equation,

y = mx + b

where 'b' = y-intercept = 0

Since, y-intercept is zero, table will represent the proportional relation between x and y.

Proportional or non proportional → Proportional

You might be interested in

caroline bought 6 pieces of gold ribbon. each piece was 3/5 of a yard long. How many yards of ribbon did she buy in all?

Zarrin [17]

Answer:

= 3 3/5 yards

Step-by-step explanation:

If 1 piece = 3/5 yard

What about 6 pieces = ?

= (6 x 3/5) ÷ 1

= 18/5 ÷ 1

= 18/5

= 3 3/5 yards

5 0

2 years ago

(7+8)+(5+9)+4(2+5) =

Inessa [10]

Answer: 57

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

If R is the midpoint of OS , RS = 2x - 4, ST = 4x – 1, and RT = 8x – 43, find QS.

RoseWind [281]

Answer:

56

Step-by-step explanation:

5 0

1 year ago

Need help badly plssssss

Ksenya-84 [330]

Answer:

translation then reflection

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

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

2 years ago