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1 year ago

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What is the inverse of the function f(x) = 2x 1? h(x) = one-halfx – one-half h(x) = one-halfx one-half h(x) = one-halfx – 2 h(x)

= one-halfx 2

1 answer:

lidiya [134]1 year ago

8 0

Answer:

214

Step-by-step explanation:

24

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How do I prove this?

Vadim26 [7]

Incorrect info please upload a correct photo of the problem

6 0

2 years ago

AP calculus HW Need help on #61

Yuki888 [10]

Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x

Crossing the 1s , the 2xs, and the x^2s out, we get

(2x</span>∆x+∆x^2-2∆x)/<span>∆x

Dividing the </span><span>∆x, we get
2x+</span><span>∆x-2.

Making the limit of </span><span>∆x=0, we get 2x-2.</span>

7 0

3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Let x = the length of

-BARSIC- [3]

gev me branly for the answer

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

HELPPP!! Plzzzz can’t do any more points so 10

Anestetic [448]

3. Answer: a) RT

b) XZ

c) TS

4. Answer: a) 26

b) 58

c) 21.5

d) 127

<u>Step-by-step explanation:</u>

2(PT) = AE 2(CP) = AN 2(CT) = EN

2(13) = AE 2(29) = AN 2(CT) = 43

26 = AE 58 = AN CT = 21.5

Perimeter = AE + AN + EN

= 26 + 58 + 43

= 127

4 0

2 years ago