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Liula [17]
3 years ago
12

8 1/3 miles 9 4/5miles 11 5/6miles 12 miles 14 7/8 miles. how many miles would that be that in a week run.

Mathematics
1 answer:
Kaylis [27]3 years ago
6 0

Answer:

8 1/3+9 4/5+ 11 5/6+12+14 7/8=<u>58 61/120</u>

Step-by-step explanation:

First, you would add all of the whole numbers: 8+9+11+1+2+14=54

Second, you would add all of the fractions 1/3+4/5+5/6+7/8=4 61/120

Lastly, 54+4 61/120=<u>58 61/120</u>

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Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,
Anton [14]

Answer:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

Step-by-step explanation:

So we have the equation:

\tan(x-y)=\frac{y}{8+x^2}

And we want to find dy/dx.

So, let's take the derivative of both sides:

\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]

Let's do each side individually.

Left Side:

We have:

\frac{d}{dx}[\tan(x-y)]

We can use the chain rule, where:

(u(v(x))'=u'(v(x))\cdot v'(x)

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])

Differentiate x like normally. Implicitly differentiate for y. This yields:

=\sec^2(x-y)(1-y')

Distribute:

=\sec^2(x-y)-y'\sec^2(x-y)

And that is our left side.

Right Side:

We have:

\frac{d}{dx}[\frac{y}{8+x^2}]

We can use the quotient rule, where:

\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}

f is y. g is (8+x²). So:

=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}

Differentiate:

=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

And that is our right side.

So, our entire equation is:

\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)

The right side cancels. Let's distribute the left:

\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2

Move -2xy to the left. So:

\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2

Factor out a y' from the right:

\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)

Divide. Therefore, dy/dx is:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}

We can factor out a (8+x²) from the denominator. So:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

And we're done!

8 0
3 years ago
Passes through(4,7)Slope=2
Anna71 [15]

Answer:

Step-by-step explanation:

Slope = 2 and passes point (4,7)

The equation is y - y1 = m(x - x1)

And y1 = 7 while x1 = 4, m = 2

: - y - 7 = 2(x - 4)

y - 7 = 2x - 8

y = 2x - 8 + 7

y = 2x - 1 this is the equation

7 0
3 years ago
I need help in the middle one
azamat
Which one is the middle one ?
4 0
3 years ago
To estimate rhe difference below,would you round the numbers to the nearest ten, nearest hundred, or nearest thousand?explain. 8
iogann1982 [59]
To the nearest thousand, because when you are estimating numbers, you always round to the nearest higher place (like the nearest tenth if comparing 26 and 18).
4 0
3 years ago
Money from an allowance or job is known as what
sineoko [7]

Answer:

salary my guesss don't know right or wrong

3 0
3 years ago
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