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2 years ago

Has anyone heard from mariahb i need her help its serious

Mathematics

1 answer:

horrorfan [7]2 years ago

6 0

Answer:

she was active 2 hrs ago

Step-by-step explanation:

sorry:(

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Complete the steps to factor 56x2 – 8x – 7x + 1 by grouping.

solmaris [256]

you dont use the distributive property

(8x-1)(7x-1)

then you foil it and combine like terms

7 0

3 years ago

Find the standard deviation for the group of data items 4, 4, 4, 4, 6, 8 The standard deviation is (Simplify your answer. Round

marysya [2.9K]

Answer: 1.53

Step-by-step explanation:

The given data : 4, 4, 4, 4, 6, 8

Number of data values : n= 6

The mean value of the given data will be :-

$\overline{x}=\dfrac{\sum_{i=1}^{6}x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{30}{6}=5$

The formula to find standard deviation:_

$\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}]$

Now,

$\sum_{i=1}^{6}(x_i-\overline{x})^2=(-1)^2+(-1)^2+(-1)^2+(-1)^2+(1)^2+(3)^2\\\\=1+1+1+1+1+9=14$

The standard deviation will be :-

$\sigma=\sqrt{\dfrac{14}{6}}\approx1.53$

5 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:

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