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2 years ago

WILL GIVE BRAINLIEST!! Need answer fast!!! 16x > - 2

Mathematics

2 answers:

zvonat [6]2 years ago

7 0

The answer is -8. Hope this helps

Alexus [3.1K]2 years ago

4 0

Answer:

x > -1/8 hope helps

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. The roots of an equ

Inessa [10]

Answer:

If the roots of an equation are x = -1 ± i, it means that the factorized form of that equation is: (x + 1 + i)(x+ 1 - i) = 0.

Using the distributive property, we have:

(x + 1 + i)(x+ 1 - i) = x^2 + x - ix + x + 1 - i + ix + i + 1

Combining like-terms and simplifying:

⇒ x^2 + x + x + 1 + 1 = x^2 + 2x + 2 = 0

Therefore, the stament is correct. If the roots of an equation are x = -1 ± i, then the equation is: x^2 + 2x + 2 = 0.

5 0

2 years ago

Help:

dalvyx [7]

Answer:

Step-by-step explanation:

a. (11x) + (22x)= 33x

b. (-6x) - (-6x) = 0

c. (-13) + (14x) - (+13t) = -13t + 14x - 13

d. (-4t) + (+4t) - (-6t) = 6t

4 0

2 years ago

Read 2 more answers

Which Property is shown?<br> (5•10) + (5•5) = 5( 10 + 5)

finlep [7]

Distributive Property

8 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

2 years ago

17.25 round up to ones

Irina18 [472]

17.24

Since 7 comes after 5 we need to round to the next number.....18.00

17.24 rounded to ones = 18.00

Hope I helped:P

7 0

2 years ago

Read 2 more answers