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Monica [59]
4 years ago
11

Select the function that has been shifted up by 2 and is less steep than the function f(x) = 3x-5

Mathematics
1 answer:
Scorpion4ik [409]4 years ago
3 0

Answer:

As the function has been shifted up by 2 and is less steep than the function f(x) = 3x-5.

Thus, option C is true. i.e. g(x)=x-3

Step-by-step explanation:

Given the function

f(x) = 3x-5

writing in the slope-intercept form

y = mx+b

where m is the line and b is the y-intercept

  • Thus, slope of f(x) = 3

To move a function up, we need to add outside the function.

f(x) + b means  f(x) moved up 'b' units.

So, by shifting up by 2 means f(x) would become g(x)= 3x-5+2 = 3x-3

But, we have to find the function which is less steep, meaning having a slope less than 3.

so, the best-suited function that has been shifted up by 2 and is less steep than the function f(x) = 3x-5 will be:

  • g (x) = x - 3

A graph is also attached.

  • In the graph, the red line represents f(x)=3x-5, and the blue line is representing the function g(x)=x-3 which has been shifted up by 2 and is less steep than the function f(x) = 3x-5.

As the function has been shifted up by 2 and is less steep than the function f(x) = 3x-5.

Thus, option C is true.

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c. x>4

Step-by-step explanation:

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Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel
nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

\theta=\frac{\pi}{3}

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be=r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j

By using cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}

Force along the plane will be=\mid F_x\mid=F\cdot r

Force along the plane will be =\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3N

By using i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0

Therefore, force along the plane=\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)

2.The vector perpendicular to the plane=r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j

The force perpendicular to the plane=\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

The force perpendicular to the plane=4N

Therefore, F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=2\sqrt 3i-6j-2\sqrt3 i-2j=-8j

Hence,sum of two component of forces=Total force.

6 0
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How do I do this? (Number 1)
german
This makes no sence number 1 what?

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