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Maurinko [17]
3 years ago
5

Graph this line using the slope and y-intercept: y=5x+2

Mathematics
2 answers:
notsponge [240]3 years ago
3 0

Answer:

You can use Desmos, or any online graph-er to graph the line.

Ber [7]3 years ago
3 0

Answer:

Step-by-step explanation:

First, plot the y-intercept, (0, 2).

Next, place your pencil there.  Move your pencil point 1 unit to the right and 5 units up.  Plot this point (1, 2+5), or (1, 7).

Draw a straight line through (0, 2) and (1, 7).

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How many real solutions does the system have?<br><br><br> {y=−3x−3y=x2−3x+5
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Find the equation of a line parallel to y=2x-4 that contains the points (5,1)
Alex73 [517]
The correct answer is
y=2x-9

Since the lines are parallel, they will gave the same slope (in this case it’s 2)

Our equation will be
y=mx+b

We know the slope is 2, and they’ve given us x and y coordinates. Plugging it in to the slope equation we get

1=2(5)+b

1=10+b

Subtract 10 from both sides and you get

-9=b

Putting it together our new parallel line will be

y = 2x - 9
5 0
3 years ago
Solve the system of equations and choose the correct ordered pair.
nasty-shy [4]
The answer is D

Because if you replace the variables just by testing them they give the answer as X=-3 and Y=3.
In which replacing:
3(-3)+5(3)=6
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6 0
3 years ago
Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec
klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

x=t^3-3t

and

y=t^2-4

The tangent function is given by:

\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }

\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}

The tangent is vertical at when \frac{dx}{dt}=0

\implies \frac{3t^2-3}{2t}=0

\implies 3t^2-3=0

\implies 3t^2=3

\implies t^2=1

\implies t=\pm1

When t=1,

x=1^3-3(1)=-2 and y=1^2-4=-3

When t=-1,

x=(-1)^3-3(-1)=2 and y=(-1)^2-4=-3

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when \frac{dy}{dx}=0 or  \frac{dy}{dt}=0

\implies 2t=0

\implies t=0

When t=0,

x=0^3-3(0)=0 and y=0^2-4=-4

There is a horizontal tangent at (0,-4)

5 0
3 years ago
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