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3 years ago

2.The coordinate plane shown below is a graph of a relation.

Mathematics

1 answer:

vazorg [7]3 years ago

5 0

I believe the answer is c

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A bucket full of milk is 80cl,how many litres are in 15 of such

PIT_PIT [208]

Answer:

1.2 liters

Step-by-step explanation:

1 bucket has a volume of 80 cL.

15 buckets have a volume of 15 * 80 cL = 120 cL

1 L = 100 cL

120 cL * (1 L)/(100 cL) = 1.2 L

5 0

3 years ago

The sum of two numbers is -18. If the first number is 10, which equation represents this situation, and what is the second numbe

ki77a [65]

Don’t know if there’s much of an equation for it, but I guess the best way to write it out is 10 + x = -18. Then, you would subtract the 10 from both sides in order to isolate the x. So then you’re left with x = -28. So -28 is your answer.

8 0

3 years ago

What is i^84? a) -i b) i c) -1 d) 1

mariarad [96]

Answer:

D) 1

Step-by-step explanation:

When you start raising i to certain powers, you begin to notice a pattern.

$i^1=i \\\\i^2=-1 \\\\i^3=-i \\\\i^4=1 \\\\i^5=i \\\\ i^6=-1 \\\\ i^7=-i \\\\i^8= 1$

This cycle repeats forever. Since 84 is a multiple of 4, i^84 must be 1. Hope this helps!

7 0

3 years ago

Factor completely and show work

sukhopar [10]

The answers to the questions

3 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 :

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