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3 years ago

Graph the linear function described by the equation y=-3x-2.

Mathematics

2 answers:

Rudik [331]3 years ago

8 0

The graph is in the picture I added. :3

Vesna [10]3 years ago

4 0

Answer:

step 1: slope= -3, y-intercept= -2

Step 2: Plot the y intercept

step 3: use the slope to plot one more point: -(0, -2)

step 4: the straight line through these points shows the graph of the function: (1, -5)

Step-by-step explanation:

EDGE 2020 test correct

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Can someone help me plz and the last two that you couldn’t see was ( c- 1 1/3 ) and ( D- 1 1/9)

prisoha [69]

Answer:

B 9/10

Step-by-step explanation:

3/5 ÷2/3

Copy dot flip

3/5 * 3/2

9/10

5 0

3 years ago

I'm a little confused, it's about Combining like terms but I did don't understand

ella [17]

2(-n-3)-7(5+2n)

First, you need to distribute the numbers in front of the parentheses to those in the parentheses. This means you multiply.

2 times -n= -2n

2 times -3= -6

-7 times 5= -35

-7 times 2n= -14n

This leaves you with

-2n -6 -35 -14n

Next, combine like terms.

(-2n) + (-14n)= -16n
(-35) + (-6)= -41

This gives you

-16n-41, which is B in your answer choices.

Hope this makes sense!! :)

6 0

3 years ago

Read 2 more answers

For what values of a are the following expressions true?/a+5/=-5-a

katovenus [111]

Explanation:

The expression is given below as

$|a+5|=-5-a$

Concept:

We will apply the bsolute rule below

$\begin{gathered} if|u|=a,a>0 \\ then,u=a,u=-a \end{gathered}$

By applying the concept, we will have

$\begin{gathered} \lvert a+5\rvert=-5-a \\ a+5=-5-a,a+5=5+a \\ a+a=-5-5,a-a=5-5 \\ 2a=-10,0=0 \\ \frac{2a}{2}=\frac{-10}{2},0=0 \\ a=-5,0=0 \end{gathered}$

Hence,

The final answer is

$a\leq-5$

5 0

1 year ago

X²+Y²=250 and XY=117<br> What are the values of X and Y?

yan [13]

Answer:

Step-by-step explanation:

$x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}$

$y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9$

6 0

4 years ago

Waves with an amplitude of 2 feet pass a dock every 30 seconds. Write an equation for a cosine function to model the height of a

kolbaska11 [484]

Answer:

The cosine function to model the height of a water particle above and below the mean water line is h = 2·cos((π/30)·t)

Step-by-step explanation:

The cosine function equation is given as follows h = d + a·cos(b(x - c))

Where:

$\left | a \right |$ = Amplitude

2·π/b = The period

c = The phase shift

d = The vertical shift

h = Height of the function

x = The time duration of motion of the wave, t

The given data are;

The amplitude $\left | a \right |$ = 2 feet

Time for the wave to pass the dock

The number of times the wave passes a point in each cycle = 2 times

Therefore;

The time for each complete cycle = 2 × 30 seconds = 60 seconds

The time for each complete cycle = Period = 2·π/b = 60

b = π/30 =

Taking the phase shift as zero, (moving wave) and the vertical shift as zero (movement about the mean water line), we have

h = 0 + 2·cos(π/30(t - 0)) = 2·cos((π/30)·t)

The cosine function is h = 2·cos((π/30)·t).

4 0

3 years ago