500 sutracts 125 is equal to?

Which method is used in elimination to find the solutions of x + 3y = 9 and x - 3y = 8?

weqwewe [10]

Answer:

x=17/2 y=1/6

Step-by-step explanation:

-1(x+3y=9)

1(x-3y=8)

_________

-x-3y=-9

x-3y=8

________

-6y/6=-1/6 y=1/6

x+3(1/6)=9

x+ 1/2=9

- 1/2 - 1/2

x=17/2

4 0

3 years ago

Suppose that you draw two cards from a deck. After drawing the first card, you do not put the first card back in the deck. What

S_A_V [24]

Answer:

(B) 0.0588

Step-by-step explanation:

The probability is calculated as a division between the number of possibilities that satisfy a condition and the number of total possibilities. Then, the probability that the first card is diamonds is:

$P_1=\frac{13}{52}$

Because the deck has 52 cards and 13 of them are diamonds.

Then, if the first card was diamonds, the probability that the second card is also diamond is:

$P_2=\frac{12}{51}$

Because now, we just have 51 cards and 12 of them are diamonds.

Therefore, the probability that both cards are diamonds is calculated as a multiplication between $P_1$ and $P_2$ . This is:

$P=\frac{13}{52}*\frac{12}{51}=\frac{1}{17}=0.0588$

8 0

3 years ago

Add 8.563 and 4.8292. then times the answer by 9.183 divide it 2 times by 1.192. Add to the answer with this number after your f

mina [271]

Answer:

105.876373581

Step-by-step explanation:

8.563 + 4.8292 = 13.3922

13.3922 × 9.183 = 122.9805726

122.9805726 ÷ 1.192 = 103.171621309

103.171621309 ÷ 1.192 = 86.5533735811

86.5533735811 + 19.323 = 105.876373581

6 0

3 years ago

Read 2 more answers

If + 3 = 27 − 3 = 9,ℎ = ? A. 3 B. 9 C. 14 D. 18 E. 36

tekilochka [14]

D you have to substitute the h with 27

Hope this helps

7 0

2 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