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3 years ago

the review words are, addend, addition, difference, number sentences, related facts, subtraction, sum

Mathematics

1 answer:

prisoha [69]3 years ago

8 0

What's the question? tell me what you need to do with the words and i can help you.

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You have $50 in your bank account.

kap26 [50]

Answer:

10 hours

Step-by-step explanation:

If I have $50 in my bank account, and I want to have a total of $130 in my account. It means that I need to work enough hours to make $130 - $50 = $80.

If I make $8 per hour, and I need to make $80, then I just have to work 10 hours. ($80/8 = 10)

7 0

3 years ago

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

3 years ago

Given g(x) = :-2 +4, solve for a when g(x) = 0.

evablogger [386]

Answer:

-7/4

Step-by-step explanation:

You are looking for the composite g(f(2)). The simplest way to solve this is to evaluate f(2) and enter the solution in to your g function.

g(f(2))=g(-(2)^2-2(2)+4)=g(-4-4+4)=g(-4)

g(-4)=4/(-4(-4)-2)=4/(16-2)=4/14=2/7

Therfor, g(f(2))=2/7 **I'm assuming the -4x-2 is all in the denominator of the g(x) function. If -2 is not in the denominator you would have

g(f(2))=4/(-4(-4)) -2=4/16 -2=1/4 -2=1/4-8/4= -7/4

8 0

3 years ago

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9

mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$