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2 years ago

Find the missing side. Can someone help???

Mathematics

1 answer:

tino4ka555 [31]2 years ago

4 0

Answer:

its not loading?

Step-by-step explanation:

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Find the value of y...

soldier1979 [14.2K]

Answer:

y=0

Step-by-step explanation:

Rewrite Evaluate Powers:

3 $3^{2y-1}+3^{-1}+2*3^{y}*3^{-1}=1$

Calculate:

$(3^{y})^{2}*\frac{1}{3} +2*3^{y}*\frac{1}{3}=1$

Solve using substitution:

$(3^{y})^{2}*\frac{1}{3} +\frac{2}{3} *3^{y}=1$ $t=3^{y}$

Solve the equation for t:

$t^{2}*\frac{1}{3}+\frac{2}{3}t=1$

t=-3

t=1

Substitute back to t=3^y

3^y=-3

3^y=1

y∉R

$3^{y}=1$

y=0

5 0

2 years ago

Can some one help me to fi d out the slope

Molodets [167]

Well the slope formula is y² - y¹ / x² - x¹.

If there was a picture I could help!

3 0

2 years ago

Read 2 more answers

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:

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