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3 years ago

How do you solve an inequality using an absolute value? Example: |x+3|<2

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

4 0

Answer:

x < -1

Step-by-step explanation:

hope that will help you

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Find the area of the composite figure.

elena55 [62]

I got area=41.7 m

Step by step:
Find A; 6x4 =24
Find B; you have to subtract 2 from 3.8 to find the length of the dashed line. 1.8x3.5=6.3
Find C; 3.8x3=11.4.
Add together; 11.4+6.3+27=41.7

8 0

3 years ago

Solve log(base 2) (x+1) + log (base 2) (x-5)= 4

Afina-wow [57]

(x+1)(x-5)= 2^4=16

x^2 -4x -5 =16

x^2-4x-21=0

x= -3 or 7
but we would choose 7 as the answer as -3 gives negative value at (x+1) and we know that log of negative number is not defined.

6 0

3 years ago

Find 3 different possibile pairs of values for a and b

Lynna [10]

a=1, b =18

a=2, b = 9

a = 3 , b = 6

3 0

3 years ago

Read 2 more answers

If angle ABC is one degree less than three times angle ABD and angle DBC=47 find each measure

Contact [7]

It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.

Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC

⇒ Angle ABC = Angle ABD + Angle DBC

Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree

Let angle ABD be x

⇒ Angle ABC = 3x-1

Also, Angle ABC = Angle ABD + Angle DBC

Substituting the values in the above equations

⇒ 3x-1 = x+47

⇒ 2x = 48

⇒ x = 24

So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree

8 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:

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

This gives the infinite many solution.

5 0

3 years ago