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3 years ago

7

Rectangle???ABCD has vertices A(8, 5) , B(8, 10) , C(14, 10) , and D(14, 5) . A dilation with a scale factor of 1.2 and centered

at the origin is applied to the rectangle. Which vertex in the dilated image has coordinates of (9.6, 6) ?
A???

B???

C???

D???

1 answer:

VLD [36.1K]3 years ago

4 0

I just took the test and its A :)

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Find the distance speed=96km/h time=20 mins

Alex_Xolod [135]

Answer:

32km

Step-by-step explanation:

speed=96km/h

time=20mins=1/3hour

<h3>Distance=Speed×Time</h3>

➜96×1/3

➜96/3

➜32km

6 0

2 years ago

Can someone please show me how to divide 22 and 121 please and take your time explaining so that way I can understand it. Thank

Aleonysh [2.5K]

When dividing 22 into 121 , you are cosidering how many times 22 goes into 121. This is called division.

Now how do we use division to find this?

well we do 22 over 121 or 22/121

The next step is the conversion to decimal.

How do we figure this out? well we use multiplication through the remainder theorem. The way this works is we figure out how many times 121 goes into a quantity of another number so,

121 goes into 22 how many times. Well 0, thus by the reaminder theorem we will add a zero assuming that we put a decimal place so now we know the number is atleast .XX

Now we have 220, so how many times does 121 go into 220? Well 1. We can add that to our number .1XX and we subtract that from 220 and then add a zero.

220-121 is 99 plus add a zero so we now have 990. 121 goes into 990 how many times? 8 times so we know the number is .18

This process could be continued forever, but there is no need. I hope I explained this well. Explaining core concepts like this generally should be done in person with someone.

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3 years ago

As part of the quality-control program for a catalyst manufacturing line, the raw materials (alumina and a binder) are tested fo

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2 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

Can someone please explain with answer?

ira [324]

Using: A^2+b^2=c^2

A&b= (√10)^2=10

10+10=c^2

20=c^2

(take the square root of 20)

c=2√5

Your answer is D

7 0

3 years ago

Read 2 more answers