Ask question

Ask question

All categories

2 years ago

6

Graph each linear equation y equals negative 3x - 2

2 answers:

den301095 [7]2 years ago

4 0

Answer: y=-(3x-2)= -3x+2

The intercept with y is 2

The intercept with x is 2/3

emmasim [6.3K]2 years ago

4 0

View the attached image for the graph.

You might be interested in

In the diagram below. C is an equal distance to A and B.

Fynjy0 [20]

Answer:

C(50;-20)

Step-by-step explanation:

The question is asking for the midpoint of the line.

The triangle is formed by placing a line directly from the y-axis to the x-axis.

The midpoint theorem is[ (x1+x2)/2 ; (y1+y2)/2 ].

your coordinates at A is (0;-40), because the x value is 0 on the line as shown in the diagram above point A.

Your coordinates for B are (100;0)

C[(x1+x2)/2 ; (y1+y2)/2 ]

C [(0+100)/2 ; (-40+0)/2 ]

C[ 100/2 ; -40/2]

C (50 ; -20)

7 0

2 years ago

Arc CD is One-fourth of the circumference of a circle. What is the radian measure of the central angle?

Sveta_85 [38]

Answer:

Answer is b

Step-by-step explanation:

just did it

8 0

2 years ago

What is the answer to this 9(-2)=

cestrela7 [59]

The answer is -18

9 · -2 = -<span>18
</span>

6 0

2 years ago

Read 2 more answers

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

astra-53 [7]

Answer:

$3\pi \rightarrow y=2\cos \dfrac{2x}{3}\\ \\\dfrac{2\pi }{3}\rightarrow y=6\sin 3x\\ \\\dfrac{\pi }{3}\rightarrow y=-3\tan 3x\\ \\10\pi \rightarrow y=-\dfrac{2}{3}\sec \dfrac{x}{5}$

Step-by-step explanation:

The period of the functions $y=a\cos(bx+c)$ , $y=a\sin(bx+c)$ , $y=a\sec (bx+c)$ or $y=a\csc(bx+c)$ can be calculated as

$T=\dfrac{2\pi}{b}$

The period of the functions $y=a\tan(bx+c)$ or $y=a\cot(bx+c)$ can be calculated as

$T=\dfrac{\pi}{b}$

A. The period of the function $y=-3\tan 3x$ is

$T=\dfrac{\pi}{3}$

B. The period of the function $y=6\sin 3x$ is

$T=\dfrac{2\pi}{3}$

C. The period of the function $y=-4\cot \dfrac{x}{4}$ is

$T=\dfrac{\pi}{\frac{1}{4}}=4\pi$

D. The period of the function $y=2\cos \dfrac{2x}{3}$ is

$T=\dfrac{2\pi}{\frac{2}{3}}=3\pi$

E. The period of the function $y=-\dfrac{2}{3}\sec \dfrac{x}{5}$ is

$T=\dfrac{2\pi}{\frac{1}{5}}=10\pi$

5 0

3 years ago

Solve for g.<br><br> g/2 = 3<br><br> g = ?

KATRIN_1 [288]

Answer:

6

Step-by-step explanation:

4 0

2 years ago