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2 years ago

How do you do surface area and volume?

2 answers:

5 0

Depending on what shape or figure you are talking about, generally volume= Bh B= area of the base and for area greatly depend on what the object is.
square= side cubed
rectangle= L time H
triangle= base time height divided by 2
circle= pi time radius cubed
that should get you through what you need.

marissa [1.9K]2 years ago

4 0

Prism; Surface area = Lateral Area + 2(base)
Volume = Base times height
Cylinder; surface area = lateral area + 2(base)
Volume = Base times height
Pyramid; Surface area = Lateral times base
Volume = 1/3 base times height
Cone; Surface area= Lateral area + base
Volume= 1/3 base times height

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sometime after passing a truck stop 300 miles from the start of her trip Cora's hybrid car ran out of gas assuming that the tank

Scrat [10]

Answer: 7.5 gallons

Step-by-step explanation:

300 divided by 40 is 7.5

Mark me as brainliest if this helps!

8 0

3 years ago

Solve the system by substitution. Check your solution.

Zepler [3.9K]

Answer:

a. (15, 15)

Step-by-step explanation:

We start with those two equations:

1) a - 1.2b = -3

2) 0.2b + 0.6a = 12

We'll begin by modifying equation #1 to isolate a:

a = -3 + 1.2b

Then we'll use this value for a in the second equation:

0.2b + 0.6 (-3 + 1.2b) = 12

0.2b - 1.8 + 0.72b = 12

0.92b = 13.8

b = 15

Then we'll place that value of b in the first equation to find a:

a - 1.2 (15) = -3

a - 18 = -3

a = 15

3 0

2 years ago

The equation y= mx+b what's the b in the equation

andreyandreev [35.5K]

Answer:

The b is the y-intercept.

the m is the slope.

6 0

2 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 :

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