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Verdich [7]
3 years ago
5

What sequence of rigid transformations takes figure C to figure D?

Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer:

D

Step-by-step explanation:

Hope this helps :)

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John is 3 years older than Jim. Jim is 4 years less than half of Dana’s age. How old is each person if their ages add up to 167?
kogti [31]

Solution:

we are given that

John is 3 years older than Jim. Jim is 4 years less than half of Dana’s age.

Let the age of Jim be x years then we can say

Age of John=x+3

If diana Age is D then

x=D/2-4

It can be re-wriiten as

Diana age D=2x+8

Given that  their ages add up to 167

x+x+3+2x+8=167\\
\\
4x+11=167\\
\\
4x=167-11\\
\\
4x=156\\
\\
x=39\\

Hence jim is 39 nyears old, John is 39+3=42 years old. and Diana is 2*39+8=78+8=86 years old.

3 0
3 years ago
Can you guys help me? I need to study this for a test 1/2=2/3x
Aleksandr [31]
1/2=2/3x
x=3/4
please brainliest
hope this help
6 0
3 years ago
I need help on rhis pls fastttttttt fastttttttt pls stytyt<br> y
zavuch27 [327]

Answer:

60 slices

Step-by-step explanation:

Well, to start you must figure the amount of slices, to begin with. This is 9 x 8

this is = to 72.

Then you take 5/6 of 72/1 by multiplying across. to get 60

7 0
3 years ago
Read 2 more answers
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} &#10;
substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
Please help me with these math problems! I have had so much going on lately and haven't had time to do them
Anvisha [2.4K]
Tanx =60/70
For more explanation
....

8 0
3 years ago
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