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2 years ago

the graph of a quadratic function f passes through the origin and the maximum value of f is 5 when x=4 find f

Mathematics

1 answer:

Hitman42 [59]2 years ago

7 0

Answer:(-5/16)x^2 +(5/2)x

Step-by-step explanation:

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Which number line shows -4 , 1,1 and 2 correctly plotted

Colt1911 [192]

Answer:

Step-by-step explanation:

The number line in option C contains the points -4, -1, 1 and 2 correctly plotted.

6 0

2 years ago

Please Help me Its a math problem see image attached

SSSSS [86.1K]

Answer:

C. - 3

Step-by-step explanation:

Average rate of change within an interval is the slope of the endpoints.

The endpoints are:

(- 1, 6) and (0, 3)

The slope is

(3 - 6)/(0 - (-1)) = - 3/1 = - 3

Correct choice is C

4 0

1 year ago

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Determine the probability of randomly selecting two individuals who are issued exactly two credit cards. [Hint: Are the events

pishuonlain [190]

Answer:

They are dependent because we have to select from people who are given cards.

Step By Step Explanation:

So we'll take away people not given cards first den find the probability of selecting people with cards over the total number of people present .

Probability we'll be equal to = number of people with card(C) two persons/total number of people

Where C represent combination

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3 years ago

Does a plane contains at least two non-collinear points.

Ganezh [65]

There could be collinear but need to be on the same plane

8 0

3 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

2 years ago

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