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3 years ago

Which best describes the error in finding the surface area of the prism?

1 answer:

3 0

To find the surface area of a prism (or any other geometric solid) we open the solid like a carton box and flatten it out to find all included geometric forms. To find the volume of a prism (it doesn't matter if it is rectangular or triangular) we multiply the area of the base, called the base area B, by the height h.

This info should help you.

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Answer to this math question

gladu [14]

Answer:

$4\frac{2}{3}$

Step-by-step explanation:

This is a recurring decimal.

Let x = 4.666666... ⇒ (1)

10x = 46.66666... ⇒ (2)

Subtract equation (1) from equation (2)

10x = 46.66666..

- x = 4.666666...

9x = 42.0000.....

9x = 42

x = 42/9

= 14/3

= $4\frac{2}{3}$

5 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago

The formula s=12at ^2 is used to calculate the distance s travelled by a bike when a=3 and t=10, each correct to the nearest int

seraphim [82]

Answer:

Step-by-step explanation:

Given the formula for calculating the distance travelled expressed as;

s=1/2at^2

Given

a = 3

t = 10

Required

lower bound of s

Substitute the given values into the equation;

s=1/2at^2

S = 1/2(3)(10)^2

S = 1/2 * 3 * 100

S = 3 * 50

S = 150

Hence the lower bound of distance S is 150

8 0

2 years ago

I think of a number and multiply it by 4. I then add 8 and I'm left with an answer of 30. What number did I think of? Form and s

postnew [5]

Answer: number= 5.5; equation: 4x+8=30

Step-by-step explanation:

Take x for the unknown number. Then just follow the words and turn them into mathematical symbols. Easy.

5 0

3 years ago

Theresa is comparing the graphs of y=2x and y=5x. Which statement is true?

abruzzese [7]

Answer:

Both graphs have a “y-intercept” at (0,0), so i believe the answer would be they don’t have y intercepts

Step-by-step explanation:

7 0

3 years ago