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3 years ago

PLZ answer fast 100 points

Mathematics

2 answers:

Kipish [7]3 years ago

5 0

Answer:

y = f(x − 2) + 4

Shifts 2 units right, 4 units up.

y = −f(x − 3)

Shifts 3 units right, reflects in x-axis.

y = f(x + 5)

Shifts 5 units left

Step-by-step explanation:

Let the initial function be y = f(x)

Then,

y = f(x − 2) + 4 is obtained by translating the function 2 units towards right and 4 units up

y = -f(x - 3) is obtained by translating the graph 3 units towards right and then reflection along the x-axis

y = f(x + 2) is obtained by translating the graph 2 units towards left

Diano4ka-milaya [45]3 years ago

3 0

Answer:

Can i have brainliest

The second one

The third one

and the last ones

Step-by-step explanation:

The x values are opposite

and the y stay the same

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I don't know who needs to hear this but I wanted to hopefully reach out to someone.

Alex787 [66]

Answer:

aw thank you sm <3 !!! needed to hear that

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Describe and correct the error in determining wheather (8,11) is a solution of y-x=3

Burka [1]

(8, 11) is the correct solution. x = 8, y = 11

Plug it in:

11-8=3

That's correct.

7 0

3 years ago

Read 2 more answers

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

Which value is the solution to the equation -8 = 6 - 7x? <br>A. -7 <br>B. - 2 <br>C. 2 <br>D. 5

r-ruslan [8.4K]

Step-by-step explanation:

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