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valkas [14]
3 years ago
5

I will give crown to the best answe PLEASE HELP ME!!!!!

Mathematics
2 answers:
Goshia [24]3 years ago
6 0
Comment
You can't get to quadrant 4 to quad 2 without 2 steps. The answers tell you that. The question is, did you use a reflection and a y or x translations or 2 reflections or 2 y or 2x translations.

Argument
You can take one point and see if it translate from beginning (purple triangle) to (yellow triangle) end. If two conditions will get you the right answer, then you have to try another point to break the tie.

We'll use Point C.
Choice A: If you reflect C over the y axis, you will go from (-5,1) to (5,1)
                 If you move C six units down, you will go from (5,1) to  (5,-5) which is where C' is.
                 That is pretty much the answer.

I could confirm it with another point which is the way your should do it.

Try Point A

Choice A:  Point A reflects across the y axis going from (-3,4) to (3,4)
                 Point A  translates down by moving from (3,4) to (3,-2) to become A'

Answer reflection and translation  in condition A

More Comment
I should do one more for you to show you that it is wrong.
You can try Choice B of the multiple Choice answers.

We will use Point C 
If we translate C across the x axis it will go from (-5,1) to (-5,-1)
The we are to translate 1 unit up. We will go from (-5,0) from (-5,-1) That's nowheres near where C' is. Choice B is wrong.

Anastasy [175]3 years ago
5 0

Answer:

The answer is A

Step-by-step explanation:

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75 = 4(2)^x
Valentin [98]

Answer:

4.229

Step-by-step explanation:

75 = 4(2)^x

75/4 = 2^x

Apply ln both sides

ln(75/4) = ln(2^x)

ln(75/4) = x ln(2)

x = ln(75/4) ÷ ln(2)

x = 4.22881869

3 0
3 years ago
Does the set {t, t Int} form a fundamental set of solutions for t^2y" -- ty' +y = 0?
Ivanshal [37]

Answer:

yes

Step-by-step explanation:

We are given that a Cauchy Euler's equation

t^2y''-ty'+y=0 where t is not equal to zero

We are given that two solutions of given Cauchy Euler's equation are t,t ln t

We have to find  the solutions are independent or dependent.

To find  the solutions are independent or dependent we use wronskain

w(x)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}

If wrosnkian is not equal to zero then solutions are dependent and if wronskian is zero then the set of solution is independent.

Let y_1=t,y_2=t ln t

y'_1=1,y'_2=lnt+1

w(x)=\begin{vmatrix}t&t lnt\\1&lnt+1\end{vmatrix}

w(x)=t(lnt+1)-tlnt=tlnt+t-tlnt=t where t is not equal to zero.

Hence,the wronskian  is not equal to zero .Therefore, the set of solutions is independent.

Hence, the set {t , tln t} form a fundamental set of solutions for given equation.

6 0
3 years ago
What is the square root of 4.84
dybincka [34]

Answer:

2.2. Hope it helps :D

6 0
3 years ago
Read 2 more answers
jayden lost 12 pounds in 3 weeks after that his lost rate was half if he lost a total of 84 pounds how many weeks did it take
Marina CMI [18]

Answer:

Jalen lost 12 pounds in the first 3 weeks

12/3=4 per week

than the weight loss slowed by half, meaning he lost 6 pounds in 3 weeks

6/3=2 per week

84-12=72 to loose after 3 weeks

72/2=36 weeks

to loose 84 pounds you need 3+36=39 week

pls mark me as brainlist

7 0
3 years ago
What is the answer to this problem<br> 2(6y-2)-3y=2
tester [92]

Answer:

y=9/6 (3/2 simplified)

Step-by-step explanation:

2(6y-2)-3y=2

12y-4-3y=2

ad 4 both side

12y-3y=6

9y=6

9/6

5 0
3 years ago
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