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3 years ago

A table measures 3.6 feet long. How long does the table measure in inches and in yards?

Mathematics

1 answer:

sattari [20]3 years ago

7 0

Answer:

43.2 inches is 3.6ft

1.2 yards is 3.6ft

You might be interested in

How do I find the percent

sleet_krkn [62]

Answer:

if you want to find a percent first you have to find the fraction. for an example for other, if 75 people out of 300 people are working you would divide 75 by 300. that would give you .25. then you multipluy by a hundred. that would give you 25 percent

Step-by-step explanation:

3 0

4 years ago

A line segment has endpoints A(4, 8) and B(2, 10). The point M is the midpoint of AB. What is an equation of a line perpendicula

Vera_Pavlovna [14]

I)

The Midpoint M of the line segment AB is found using the Midpoint formula:

$M=( \frac{4+2}{2} , \frac{8+10}{2})=(3, 9)$

ii)

the slope of the line through A and B is found by the slope formula:

$m= \frac{y_1-y_2}{x_2-x_1}= \frac{10-8}{2-4}= \frac{2}{-2}=-1$

iii)

the product of the slopes of 2 perpendicular lines is -1, so the slope of the line perpendicular to the line through A and B is -1/(-1)=1

iv)

the equation of the line with slope 1, which contains point M(3, 9) is found by the slope point form equation of a line:

y-9=1(x-3)

y-9=x-3

y=x+6

Answer: y=x+6

7 0

3 years ago

(d). Use an appropriate technique to find the derivative of the following functions:

natima [27]

(i) I would first suggest writing this function as a product of the functions,

$\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x$

then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.

We then have, using the power and chain rules,

$\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}$

$\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}$

For the third function, we first rewrite in terms of the logarithmic and the exponential functions,

$h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}$

Then by the chain rule,

$\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x$

By the product rule, we have

$\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}$

You could simplify this further if you like.

In Mathematica, you can confirm this by running

D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]

The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in

FullSimplify[%]

(% refers to the last output)

If it still doesn't match, you can try running

Reduce[<our result> == %, {}]

and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)

(ii) Given

$\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}$

differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives

$\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)$

$\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}$

$\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}$

which could be simplified further if you wish.

In M, off the top of my head I would suggest verifying this solution by

Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]

but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,

ResourceFunction["ImplicitD"][<our equation>, x]

but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.

3 0

2 years ago

Use the linear approximation (1+x)^kapprox 1+kx to find an approximation for the function f(x) for values of x near zero.

padilas [110]

A) approx = 1-8x
b) f(x) -8(1-x)^-1 approx = -8(1+x) = -8 - 8x
because 1/x = x^-1...

c) f(x) = (1+x)^-1/2    approx = 1- 1/2x
d) f(x) = (4+x^2)^1/2   approx = 4 + 1/2x^2
e) f(x) = (6+3x)^1/3   approx = 6+x

8 0

3 years ago

(-3,0),(5,-2) find the slope

Mashcka [7]

Answer:

-4

Step-by-step explanation:

Point slope form: y - y = m (x - x)

5 - (-3) = m (-2 - 0)

5 + 3 = m (-2)

8 = m (-2)

m = 8/(-2)

m=4

3 0

3 years ago