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svetoff [14.1K]
3 years ago
12

What value of a in the equation ax − 4 = 2y would give a line with a slope of 2?

Mathematics
1 answer:
sukhopar [10]3 years ago
3 0

Answer:

a=4

Step-by-step explanation:

ax − 4 = 2y

We need to get the equation is slope intercept form  y= mx+b.

Lets solve for y

Divide each side by 2

a/2 x - 4/2 = 2y/2

a/2 x -2 = y

or y=a/2 x -2

The slope is a/2 and the y intercept is -2

We want a slope of 2  so a/2 =2

a/2=2

Multiply each side by 2

a/2 * 2 =2*2

a =4

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3 0
2 years ago
Hi guys can someone please help me with these questions please
Zinaida [17]

Hi There!

----------------------------------

Question 2:

Obviously the right equation for this question is Point Slope Form.

Point Slope Form: y - y1 = m(x - x1)

y1 = -1

x1 = 3

m = 2

Question 2 Answer: y + 1 = 2(x - 3)

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Question 3:

Again use point slope form.

Find Slope:

7 - 1 = 6

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Question 3 Answer: y - 1 = -2(x - 1)

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Hope This Helps :)

6 0
3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term 2x^2, which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form  f(x)=ax^2+bx+c is given by the expression:

x_v=\frac{-b}{2a}

Since in our case a=2 and b=-3, we get that the x-position of the vertex is:

x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6

See the graph produced in the attached image.

4 0
2 years ago
Line<br> Passing through (4,5) and perpendicular to <br> Equation y= -2x+3
DENIUS [597]

Answer:

1/2x+3

Step-by-step explanation:

I graphed it

3 0
3 years ago
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