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2 years ago

The price of a dvd is $24.00 plus 8% sales tax

Mathematics

1 answer:

hram777 [196]2 years ago

5 0

Answer:

25.92

Step-by-step explanation:

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Use the graph to determine the domain and range of the relation, and whether the relation is a function

ohaa [14]

The answer is C. The domain is all points between -8 and 8 since it keeps waving back and forth. The range is all real numbers the graph continues on forever. And it is not a function because it does not pass the vertical line test

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3 years ago

at Phil's work there are 12 part time workers and 18 full time workers what is the ratio of part time workers to total workers

BabaBlast [244]

Answer:

the ratio of part time workers to total workers are 12/30

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3 years ago

Here are two steps from the derivation of the quadratic formula,

BaLLatris [955]

Answer:

Step-by-step explanation:

Taking the square root of both sides.

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3 years ago

By how much does the dependent variable change in response to a change of 1 unit of the independent variable?

dalvyx [7]

Answer: C) 5

--------------

x = independent variable, y = dependent variable

Assuming this is a linear function, each increase of x by 2 leads to y going up by 10. So 10/2 = 5 is the unit increase each time x bumps up by 1.

-------------------

An alternative is to use the slope formula to get

m = (y2 - y1)/(x2 - x1)

m = (25 - 15)/(4 - 2)

m = 10/2 <--- this expression shows up again

m = 5 <---- leading to the same answer as before

So we see that the slope formula is a more drawn out method to finding the answer.

6 0

3 years ago

Read 2 more answers

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago