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IRINA_888 [86]
3 years ago
14

End behavior of f(x)=2x

Mathematics
1 answer:
Alla [95]3 years ago
7 0
The answer is f(2) simply insert the x
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2(x-3)-8< -4 I need a explanation and the answer please.
tresset_1 [31]

Answer:

The solution to the inequality is:

x

The number line solution graph is also attached below.

Step-by-step explanation:

Given the inequality

2(x-3)-8< -4

Add 8 to both sides

2\left(x-3\right)-8+8

2\left(x-3\right)

Divide both sides by 2

\frac{2\left(x-3\right)}{2}

x-3

Add 3 to both sides

x-3+3

x

Thus, the solution to the inequality is:

x

The number line solution graph is also attached below.

6 0
2 years ago
If I’m 24 in 1979 what year was I born in?
Elan Coil [88]

Answer:

1950

Step-by-step explanation:

If you were 24 in 1974, simply subtract 24 from 1974 to find what year you were born in:

1974-24= 1950

Of course, it may be +/- 1 depending on what month you were born in, and what month it is now!

<em>I hope this helps! :)</em>

8 0
3 years ago
In a direct variation, the constant of proportionality is -3. Which of the following would be its equation?
saw5 [17]

Answer:

C

Step-by-step explanation:

The equation of direct variation is y=kx.

K is the constant of proportionality, which is in this case is -3.

y=-3x

7 0
3 years ago
Elm Street is straight. Willard's house is at point H between the school at point S and
Vlad1618 [11]

Answer:

7.5 Miles.

Step-by-step explanation:

3 + 4.5 = 7.5 Miles.

4 0
2 years ago
(a) Find parametric equations for the line through (1, 4, 4) that is perpendicular to the plane x − y + 2z = 7. (Use the paramet
Alex777 [14]

Answer:

x=1+t\\y=4-t\\z=4+2t

Step-by-step explanation:

A vector perpendicular to the plane ax+by+cz+d=0 is of the form (a,b,c).

So, a vector perpendicular to the plane x − y + 2z = 7 is (1,-1,2).

The parametric equations of a line through the point (x_0,y_0,z_0) and parallel to the vector (a,b,c) are as follows:

x=x_0+at\\y=y_0+bt\\z=z_0+ct

Put (x_0,y_0,z_0)=(1,4,4) and (a,b,c)=(1,-1,2)

Therefore,

x=1+t\\y=4-t\\z=4+2t

xy-plane:

Put z = 0 ⇒ t = -2 ⇒x = - 1 , y = 6

So, at point (-1,6,0)

yz-plane:

Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2

So, at point (0,5,2)

xz-plane:

Put y = 0 ⇒ t = 4 ⇒ x = 5, z = 12

So, at point (5,0,12)

5 0
2 years ago
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