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3 years ago

PLEASE HELP PLEASE PLEASE The graph below plots a function f(x):

Mathematics

1 answer:

lara31 [8.8K]3 years ago

8 0

Average rate of change = y2-y1/ x2-x1. so pick the y intercept and 3 second points and plug in to calculate.

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( m + 4 ) ( m + 1 )

love history [14]

(m+4)(m+1)

=m^2+5m+4........

3 0

3 years ago

The drama club is selling tickets to a play for each ticket they sell the school receives x dollars and the drama club receives

bogdanovich [222]

The drama club receives $7 for each ticket that they sell.

Step-by-step explanation:

Given,

Amount received by school = x dollars

Amount received by school for 96 tickets = 96x

Amount received by drama club = remaining amount after school

Total amount of 96 tickets = 96x+672

Here,

96x is the share of school

672 is the share of drama club for 96 tickets.

Therefore;

96 tickets = dollars

1 ticket = $\frac{672}{96}$

1 ticket = 7 dollars

The drama club receives $7 for each ticket that they sell.

Keywords: division, addition

Learn more about division at:

#LearnwithBrainly

6 0

3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

F(x)=9x^2 g(x)= sqrt 12-x/3 (f o g)(-4)

Slav-nsk [51]

$f(x)=9x^2 \\ g(x)=\frac{\sqrt{12-x}}{3} \\$ So,first step is to write $(fog)(-4)) =f[g(-4)] \\$

Now we start from inner paranthesis ,we need to first find value of $g(-4) =\frac{\sqrt{12-(-4)}}{3}\\ =\frac{\sqrt{12+4}}{3}\\
=\frac{\sqrt{16}}{3}\\
=\frac{4}{3}\\
(fog)(-4)) =f[g(-4)] =f(\frac{4}{3}) =9(4/3)^{2} =9*(16/9) =144/9 =16$

8 0

3 years ago

PLEASE HELP ME WITH THIS

Alik [6]

Answer:

x + y ≤ 600
5x +7y ≥ 3500
it is possible

Step-by-step explanation:

a) We can write two inequalities, one for the number of tickets, and one for the necessary revenue.

x + y ≤ 600 . . . . . . . limit imposed by available seating

5x +7y ≥ 3500 . . . . required revenue to meet expenses

b) For x = 330, the first inequality puts one limit on y:

330 +y ≤ 600

y ≤ 270

And the second inequality puts another limit on y:

5(330) +7y ≥ 3500

7y ≥ 1850 . . . . subtract 1650

y ≥ 264.3 . . . . divide by 7

The number of tickets that must be sold to meet expenses is 265, which is less than the number that can be sold, 270. It is possible to meet expenses.

4 0

3 years ago