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3 years ago

What point is on the graph of f(x)=3x4

Mathematics

1 answer:

Artyom0805 [142]3 years ago

5 0

Answer:

There are an infinite number of points on this graph. Two of them would be (1, 3) and (0, 0).

Step-by-step explanations:

In order to find any points on the graph, simply put the x value in to get the y value.

f(x) = 3x^4

f(x) = 3(1)^4

f(x) = 3(1)

f(x) = 3

(1, 3)

And then for the second point.

f(x) = 3x^4

f(x) = 3(0)^4

f(x) = 3(0)

f(x) = 0

(0, 0)

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Find the equation of the line with Slope = 6 and passing through (5,38). Write your equation in the form y =mx+b

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Hello.

We have:

A point that the line passes through
The slope of the line

Plug the values into the Point-Slope Formula:

$\mathrm{y-y_1=m(x-x_1)}$

$\mathrm{y-38=6(x-5)}$

$\mathrm{y-38=6x-30}$

$\mathrm{y=6x-30+38}$

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An art studio charges a start-up fee to cover art supplies, plus an additional fee per class. The graph shows the total cost for

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Answer/Step-by-step explanation:

✔️The rate of change can be calculated using the coordinates of any two points on the graph. Let's use (2, 60) and (3, 75):

$rate of change = \frac{y_2 - y_1}{x_2 - x_1} = \frac{75 - 60}{1} = \frac{15}{1} = 15$

Rate of change (m) = 15

✅In the context of this situation, the rate of change can be interpreted to be the additional fee per class. Thus, an additional fee per class cost $15.

✔️Initial value is the same as the y-intercept.

To find the y-intercept (b), substitute m = 15, x = 2, and y = 60 into y = mx + b.

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60 = 15(2) + b

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2 years ago

On a coordinate plane, rhombus W X Y Z is shown. Point W is at (7, 2), point X is at (5, negative 1), point Y is at (3, 2), and

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Answer:

$P = 4\sqrt{13}$

Step-by-step explanation:

Given

$W = (7, 2)$

$X = (5, -1)$

$Y = (3, 2)$

$Z =(5, 5)$

Required

The perimeter

To do this, we first calculate the side lengths using distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2$

So, we have:

$WX = \sqrt{(5- 7)^2 + (-1 - 2)^2$

$WX = \sqrt{13}$

$XY = \sqrt{(3-5)^2 + (2--1)^2}$

$XY = \sqrt{13}$

$YZ = \sqrt{(5-3)^2 + (5-2)^2}$

$YZ = \sqrt{13}$

$ZW = \sqrt{(7 - 5)^2 + (2 - 5)^2}$

$ZW = \sqrt{13}$

The perimeter is:

$P = WX + XY + YZ + ZW$

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$P = 4\sqrt{13}$

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