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3 years ago

Ghchdhdjfhfjfjfjfjfjdjdjdjf

Mathematics

2 answers:

Artyom0805 [142]3 years ago

8 0

Answer:

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Step-by-step explanation:

butalik [34]3 years ago

6 0

Answer:

hshdjfujhhshushfui

Step-by-step explanation:

You might be interested in

Out of 45 times at bat, Raul got 19 hits. Find Raul's batting average as a decimal rounded to the nearest thousandth.

tangare [24]

19/45 is 0.42222222 if you round that to the nearest thousandth you get 0.422.

6 0

3 years ago

Analyze the diagram below and complete the statement that follows.

chubhunter [2.5K]

But where is the diagram?

5 0

3 years ago

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Match the pairs of values of f(x) and g(x) with the corresponding values of h(x) if h(x) = f(x) ÷ g(x). Tiles f(x) = x2 − 9, and

vitfil [10]

For this case we must solve each of the functions.
We have then:

f (x) = x2 - 9, and g (x) = x - 3
h (x) = (x2 - 9) / (x - 3)
h (x) = ((x-3) (x + 3)) / (x - 3)
h (x) = x + 3

f (x) = x2 - 4x + 3, and g (x) = x - 3
h (x) = (x2 - 4x + 3) / (x - 3)
h (x) = ((x-3) (x-1)) / (x - 3)
h (x) = x-1

f (x) = x2 + 4x - 5, and g (x) = x - 1
h (x) = (x2 + 4x - 5) / (x - 1)
h (x) = ((x + 5) (x-1)) / (x - 1)
h (x) = x + 5

f (x) = x2 - 16, and g (x) = x - 4
h (x) = (x2 - 16) / (x - 4)
h (x) = ((x-4) (x + 4)) / (x - 4)
h (x) = x + 4

4 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:

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

This gives the infinite many solution.

5 0

3 years ago

Can you help me solve systems of equations by elimination step by step 6x+y=15 -7x-2y=-10

Anna71 [15]

Answer: (4, -9)

Step-by-step explanation:

Use elimination method. Manipulate one (or both) equations to eliminate one of the variables and solve for the remaining variable. I will be eliminating y

6x + y = 15 → 2(6x + y = 15) → 12x + 2y = 30

-7x - 2y = -10 → 1(-7x + 2y = -10) → -7x - 2y = -10

5x = 20

x = 4

Next, replace "x" with "4" into either equation and solve for y.

6(4) + y = 15

24 + y = 15

y = -9

Check:

Plug in x = 4 and y = -9 into the other equation to verify it makes a true statement.

-7x - 2y = -10

-7(4) - 2(-9) = -10

-28 - -18 = -10

-28 + 18 = -10

-10 = -10 $\checkmark$

6 0

3 years ago