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2 years ago

A factory needs to produce 1,800 jars of peanut butter on a certain day.

Mathematics

1 answer:

finlep [7]2 years ago

7 0

Answer:

Step-by-step explanation:

because in 2 they make 600 and 600 times 3 is 1,800 then do 2 times 3 witch is 6 so 6 hours after 10:30 is 3:30. I hope that makes sense <3

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A tank is full of water. find the work w required to pump the water out of the spout. (use 9.8 for g and 3.14 for Ï. if you ente

MissTica

Is this algebra?, i can help you with thats the suject

8 0

3 years ago

X²+8x-84=0 Formúla General

Brums [2.3K]

Answer:

$x_{1} =-14\\\\x_{2} =6$

Step-by-step explanation:

$x^{2} +8x-84=0\\\\x=\dfrac{-b \pm \sqrt{b^{2}-4ac } }{2a} \\\\a=1; \: \: b=8; \: \: c=-84\\\\x=\dfrac{-8 \pm \sqrt{8^{2}-4 \cdot 1 \cdot (-84) } }{2 \cdot 1}=\dfrac{-8 \pm \sqrt{400 } }{2}=\dfrac{-8 \pm20}{2} =-4 \pm 10\\\\x_{1} =-14\\\\x_{2} =6$

8 0

3 years ago

If you miss the bus, then you will be late to school.

Tpy6a [65]

law of syllogism basically is a=b b=c then a=c

a = miss bus

c= detention

so if you miss the bus you get detention

4 0

3 years ago

Read 2 more answers

Factor completely, then place the factors in the proper location on the grid.

Lyrx [107]

4(x+5y)(x-3y)
I don't know what you mean by placing the factors on the grid

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 :

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