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1 year ago

1. Use the function rule. Find y for x = 0, 1, 2, and 3. y = 4x – 2

Mathematics

1 answer:

HACTEHA [7]1 year ago

7 0

The value of y at x = 0, 1, 2, and 3 is -2 ,1,6,10 .

The missing options are

y = –2, 2, 6, 10

y = 2, 6, 10, 14

y = 4, 6, 8, 10

y = –4, –2, 0, 2

<h3>What is a Function ?</h3>

A function is a law that defines relation between a dependent variable and an independent variable .

A function is given

y = 4x-2

x = 0,1,2,3

y = ?

At x = 0

y = -2

x = 1

y = 2

at x =2

y = 6

at x = 3

y = 10

The value of y at x = 0, 1, 2, and 3 is -2 ,1,6,10 .

To know more about Function

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Write in descending order. 360t + 10t3 - 120t2

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Answer:

30t - 120t² + 10t³

Step-by-step explanation:

30t + 10t³ - 120t²

In order of degrees of function, t³ > t² > t

So in terms of order of decreasing order, we have to rearrange the equation to

30t - 120t² + 10t³ likewise in terms if increasing degree of function,

10t³ - 120t² + 30t

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2 years ago

If FX equals 2 x 2 + 1 what is f x when x equals 3

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F(3)=2(3)^2=1 would be 19

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2 years ago

Please choose the option that would best fit the empty space above: Group of answer choices only one optimal solution multiple o

Scrat [10]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In this question, some of the data is missing, that's why this question can be defined as follows:

It Includes an objective feature coefficient, its sensitivity ratio is the ratio for values on which the current ideal approach will remain optimal.

When there is Just one perfect solution(optimal solution) then the equation is:

$Max \ Z = \$ 5x_1 + \$10x_2\\\\Subject\ to: \\\\8x_1 + 5x_2 \leq 80\\\\2x_1 + 1x_2 \leq 100\\\\x1, x2 \geq 0$

When there are Several perfect solutions then the equation is:

$Max \ Z = \$ \ 200x_1 + \$ \ 100x_2\\\\ Subject \ to:\\\\8x_1 + 5x_2 \leq 80\\\\2x_1 + 1x_2 \leq 100\\\\x1, x2 \geq 0$

When there is also no solution, since it is unlikely then the equation is:

$Max \ Z = \$40x_1 + \$10x_2 \\\\Subject \ to:\\\\8x_1 + 5x_2 \leq 80\\\\2x_1 + 1x_2 \geq 100\\\\x_1, x_2 \geq 0$

When there is no best solution since it is unbounded then the equation is:

$Max \ Z = \$ 30x_1 + \$15x_2 \\\\Subject to:\\\\8x_1 + 5x_2 \geq 80\\\\2x_1 + 1x_2 \geq 100\\\\x_1, x_2 \geq 0\\$

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2 years ago

How do you do this. Please give a full answer and explain why.

OverLord2011 [107]

Law of Exponent

$\displaystyle \large{ {a}^{ - n} = \frac{1}{ {a}^{n} } }$

Compare the terms.

$\displaystyle \large{ {a}^{ - n} = {( - 2)}^{ - 3} }$

Therefore, a = -2 and n = 3. From the law of exponent above, we receive:

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{ {( - 2)}^{ 3} } }$

Exponent Def. (For cubic)

$\displaystyle \large{ {a}^{3} = a \times a \times a }$

Factor (-2)^3 out.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{( - 2) \times ( - 2) \times ( - 2)}}$

(-2) • (-2) = 4 | Negative × Negative = Positive.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{4 \times ( - 2)}}$

4 • (-2) = -8 | Negative Multiply Positive = Negative.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{ - 8}}$

If either denominator or numerator is in negative, it is the best to write in the middle or between numerator and denominators.

Hence,

$\displaystyle \large \boxed{ {( - 2)}^{ - 3} = - \frac{1}{ 8}}$

The answer is - 1 / 8

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Answer: I forgot i remember the answer in my head I forgot..

Step-by-step explanation:

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