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3 years ago

3) Find f(3) for the function below. f(x) = 2x² – 3x

Mathematics

2 answers:

arlik [135]3 years ago

7 0

Answer:

Step-by-step explanation:

$f(x) = 2x^2 -3x\\ f(3) = ?\\\\ f(3) = 2(3)^2 -3(3)\\ f(3) = 2(9) - 9\\ f(3) = 18- 9\\ f(3) = 9$

Firdavs [7]3 years ago

5 0

Answer:

Step-by-step explanation:

f(x) = 2x² – 3x

Let x=3

f(3) = 2 * 3^2 - 3*3

= 2 *9 - 9

= 18-9

= 9

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What is y=-4/3x +6. y=2

Mekhanik [1.2K]

The given equation is:
[/tex]y=-\frac{4}{3}x+6 [/tex] and we need to evaluate this for y=2.
So, first step is to plug in 2 for y in the above equation. Therefore,

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3 years ago

If f(x) = kx^3+ x^2 − kx + 2, find a number k such that the graph of f contains the point (k, -10)

Elina [12.6K]

If we write $k$ where we see $x$ in the equation and set the result equal to $-10$ , we get the result.

$f(k)=k(k)^3+k^2-k(k)+2$
$=k^4+k^2-k^2+2$
$=k^4+2=-10$
$k^4=-12$
$k_{1}=\sqrt[4]{3}(-1-i)$
$k_{2}=\sqrt[4]{3}(-1+i)$
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1 year ago

Matters has 456:8 What s his rate

Olegator [25]

We are given ratio 456:8.

The given ratio can be written in fraction form as $\frac{456}{8}$

In order to find the rate, we need to divide 456 by 8, because it would give the unit value.

When we divide 456 by 8, we first take multiple of 8 closer to the number 45.

8*6=48 but it's greater than 45, so we would take 8*5 =40.

Subtracting 40 from 45, we get 5.

Getting 6 down, we get 56. Take multiple of 8 upto 56.

8*7= 56.

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So, on dividing 456 by 8 we got 57.

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3 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

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

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

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3 years ago

Which equation can be used to determine the distance between the origin and (-2,-4)?

loris [4]

Answer:

I believe it is the second 1

5 0

3 years ago