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3 years ago

Find the unknown values in the table below

Mathematics

2 answers:

PolarNik [594]3 years ago

5 0

Answer:

Step-by-step explanation:

Table values????

frozen [14]3 years ago

3 0

Where are the values and where is the table

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PLEASE HELP AND IF ITS WORNG I WILL BE UPSET BECAUSE THIS QUESTION IS ABOUT WILL I DIE OR LIVE

IrinaVladis [17]

Answer:

B/2

Step-by-step explanation:

Jeez better not get this wrong... anyway, to start off the total is 19 books. Adding 5 then adding 3, She returned 8 books to the 19 she checked out. Uhhhh no clear answer.

NOT the first one cause 8 isn't the total. Don't think it would be the last one cause that's no fun. So it's either the second or the third one. The third one also makes no sense cause it only includes the returns from yesterday. I'm going with the second one

6 0

3 years ago

Read 2 more answers

The drama club sold bags of candy and cookies to raise money for the spring show. Bags of candy cost $8.00, and bags of cookies

vodka [1.7K]

Answer:

Number of bags of candy = 2

Number of bags of cookies = 7

Step-by-step explanation:

Let, number of bags of candy = X

So, Number of bags of cookies = X + 5

Cost of candy bag = $8

Cost of cookies bag = $2.5

Total sales = $33.50

So, Total cost of Candy Bags + Total cost of cookies bags = Total sales

= 8X + 2.5 ( X +5) = 33.50

= 8X + 2.5X + 12.5 = 33.50

= 10.5X = 33.50 - 12.5

= 10.5X = 21

X = 2

So, number of bags of candy = 2

Number of bags of cookies = 2 + 5 = 7

4 0

3 years ago

A car travels at an average speed of 64 mph. How many miles does it travel in five hours and 45 minutes

Gwar [14]

Answer:

368 miles

Step-by-step explanation:

Sorry if its wrong

5 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 :

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