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2 years ago

What is the slope of a line that is perpendicular to y=1/2x+9 that passes through (-7,-4)

1 answer:

6 0

Hi there!

To find the perpendicular slope you need to flip the fraction and change the sign. So 1/2=2/1 and tge original slope was positive, so the slope is -2. Now you sub in the point (-7,-4) in for x and y in the formula y=mx+b and solve for b (sub in 2 for m as well)
Y=mx+b
-4=-2*-7+b
-4=14+b
-4-14=b
B=-18
The equation is y=-2x-18

Hope this helps!

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3 years ago

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

Amir pits £3035 into a bank account. The account pays 4% compound interest each year. Work out how much money amir will have in

myrzilka [38]

Answer:

$\£3840.2$

Step-by-step explanation:

Let P denotes principal amount, T denotes time period and R denotes rate of interest.

Amount = $P(1+\frac{R}{100})^T$

Amir pits £3035 into a bank account. The account pays 4% compound interest each year.

Put $P=\£3035,\,R=4\%,\,T=6$

Amount = $3035(1+\frac{4}{100})^6=3035(\frac{104}{100})^6=\£3840.2$

Therefore, Amir will have $\£3840.2$ in the account after 6 years.

7 0

3 years ago

Please help me find X

Zanzabum

Answer:

x = 2

Step-by-step explanation:

The product of distances from the intersection of secants to the near and far intersections with the circle are the same. For a tangent, the near and far points of intersection with the circle are the same. This relation tells us ...

(2√3)(2√3) = x(x +4)

12 = x² +4x

16 = x² +4x +4 . . . . . add the square of half the x-coefficient to complete the square

4² = (x +2)² . . . . . . . . write as squares

4 = x +2 . . . . . . . . . . positive square root

2 = x . . . . . . . . . . . . . subtract 2

_____

<em>Alternate solution</em>

If you believe x to be an integer, you can look for factors of 12 that differ by 4.

12 = 1×12 = 2×6 = 3×4

The factors 2 and 6 differ by 4, so x=2 and x+4=6.

6 0

3 years ago