Hello i would like help on this cause i’m not sure if i did it right

1 answer:

6 0

When the problem asks for you to "dilate" I'm pretty sure it means for you to increase or make bigger. Think about when some people's pupils dilate, kinda like that.

We need to dilate AB by a factor of 2. So we will simply multiply the coordinates by 2:

For A(-3, 1)

-3 × 2= -6 and 1 × 2= 2

It is now A(-6, 2)

For B( 1, 4) you will do the same

1 × 2= 2 and 4 × 2= 8

It is now B(2, 8)

Now you can plot these new points and you will see that segment AB has become longer. In fact, it has become two times longer than it originally was. That's because we dilated the segment by a factor of 2 ;)

( Hope this helps :D !)

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A 20 foot ladder is resting against the side of the house the base of the latter is 4 feet away from the house approximately how

notka56 [123]

The ladder resting against the house forms a triangle similar to the one in the image below

To answer this question you must use Pythagorean theorem

$a^{2} +b^{2}=c^{2}$

a and b are the legs (the sides that form a perpendicular/right angle)

c is the hypotenuse (the side opposite the right angle)

In this case...

a = x

b = 4

c = 20

^^^Plug these numbers into the theorem

$x^{2} +4^{2} =20^{2}$

solve for x

$x^{2}$ + 16 = 400

$x^{2}$ = 384

x = 8√6

x ≈ 19.5959....

Hope this helped!

~Just a girl in love with Shawn Mendes

5 0

3 years ago

Calculus: Help ASAP

wariber [46]

$\bf \displaystyle \int~\cfrac{sec^2(x)}{\sqrt{1+tan(x)}}\cdot dx\\\\
-------------------------------\\\\
u=1+tan(x)\implies \cfrac{du}{dx}=sec^2(x)\implies \cfrac{du}{sec^2(x)}=dx\\\\
-------------------------------\\\\
\displaystyle \int~\cfrac{\underline{sec^2(x)}}{\sqrt{u}}\cdot\cfrac{du}{\underline{sec^2(x)}}\implies \int~\cfrac{1}{\sqrt{u}}\cdot du\implies \int~u^{-\frac{1}{2}}\cdot du
\\\\\\
2u^{\frac{1}{2}}\implies 2\sqrt{1+tan(x)}+C$