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3 years ago

What are the ordered pairs of f(x)=3-2x

Mathematics

2 answers:

choli [55]3 years ago

6 0

Answer:

y intercept: (0,3)

x intercept: (3/2,0)

Step-by-step explanation:

X intercept:

0=3-2x

2x=3

x=3/2

Y intercept:

f(0)= 3-2(0)

=3-0

soldier1979 [14.2K]3 years ago

4 0

Answer:

(0, 3) , (1, 1) , (-2, -1). Hope this is right!

Step-by-step explanation:

You might be interested in

15 is what percent of 20

Lera25 [3.4K]

100%/x%=20/15
(100/X)*x=20/15*x --I multiply both sides of the equation by X
100=1.33333333333*x --I divided both sides of the equation by .........................................(1.33333333333) to get X
100/1.33333333333=x
75=x
x=75

15 is 75% of 20

6 0

3 years ago

(a) The parametric equations x = f(t) and y = g(t) give the coordinates of a point (x, y) = (f(t), g(t)) for appropriate values

Gnesinka [82]

Answer:

Step-by-step explanation:

a). Given a parametric equation, we are describing a set of coordinates based on the value of t. The variable t is called the parameter.

b) we have the following equations. x=t y=t^2, so in order for us to know where the object is at t=t' we must replace t with the specific value t'. Hence, when t=0 the object is at (0,0^2) = (0,0) (the origin). When t=6, the object is at (6,6^2) = (6,36).

c). To eliminate the parameter, we replace the parameter in one equation by using the second equation. Recall that we have that x=t. Then, by replacing in the second equation, we have the following

$y=t^2 = (x)^2 = x^2$

where $x\geq 0$

5 0

3 years ago

A radio telescope has a parabolic surface, as shown below.

krek1111 [17]

OK, so the graph is a parabola, with points x=0,y=0; x=6,y=-9; and x=12,y=0

Because the roots of the equation are 0 and 12, we know the formula is therefore of the form

y = ax(x - 12), for some a

So put in x = 6

-9 = 6a(-6)

9 = 36a

a = 1/4

So the parabola has a curve y = x(x-12) / 4, which can also be written y = 0.25x² - 3x

The gradient of this is dy/dx = 0.5x - 3

The key property of a parabolic dish is that it focuses radio waves travelling parallel to the y axis to a single point. So we should arrive at the same focal point no matter what point we chose to look at. So we can pick any point we like - e.g. the point x = 4, y = -8

Gradient of the parabolic mirror at x = 4 is -1

So the gradient of the normal to the mirror at x = 4 is therefore 1.

Radio waves initially travelling vertically downwards are reflected about the normal - which has a gradient of 1, so they're reflected so that they are travelling horizontally. So they arrive parallel to the y axis, and leave parallel to the x axis.

So the focal point is at y = -8, i.e. 1 metre above the back of the dish.

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Csqrt%7B12%7D%20%5Ctimes%20%5Csqrt%7B12%7D" id="TexFormula1" title="\sf \sqrt{12} \t

Darina [25.2K]

Answer:

$12$

Step-by-step explanation:

$\sqrt{12} \times\sqrt{12}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{a}=a,\:\quad \:a\ge 0$

$\sqrt{12}\sqrt{12}=12$

$=12$

7 0

3 years ago

What is the distance to the earth’s horizon from point P?<br><br> Express your answer as a decimal

rewona [7]

Answer:

The distance to the earth's horizon from point P is 216.2198187 mi, appriximately 216.22 mi

Step-by-step explanation:

This is a right triangle:

Hypotenuse: c=3959 mi + 5.9 mi → c=3964.9 mi

Leg 1: a=x=?

Leg 2: b=3959 mi

Using the Pytagorean theorem:

a^2+b^2=c^2

Replacing the known values:

(x)^2+(3959 mi)^2=(3964.9 mi)^2

Solving for x: Squaring:

x^2+15,673,681 mi^2=15,720,432.01 mi^2

Subtracting 15,673,681 mi^2 both sides of the equation:

x^2+15,673,681 mi^2-15,673,681 mi^2=15,720,432.01 mi^2-15,673,681 mi^2

x^2=46,751.01 mi^2

Square root both sides of the equation:

sqrt(x^2)=sqrt(46,751.01 mi^2)

x=216.2198187 mi

x=216.22 mi

3 0

3 years ago