4x-3=2x+7 <br>algebraic form

k0ka [10]

25 lbs pounds = 400 ounces so 25 pounds is greater then350 ounces

7 0

3 years ago

9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will rema

DerKrebs [107]

Answer:

1 Shirt and 185cm remaining cloth

Step-by-step explanation:

2m and 15cm = 215cm

40m = 400cm

400/215=1.860465116

A shirt cant be in decimal so we remove the decimals, and take the as the answer. For the decimals, they are the remaining cloth.. to calculate the amount of remaining cloth we do:

1.860465116-1

=860465116

215*860465116=185cm

5 0

2 years ago

Need help with the question, I already have the degree and leading coefficient but not the constant term for this polynomial. f(

melisa1 [442]

Degree: 7
term: -3x^7
coefficient: -3

3 0

3 years ago

Solve for x 4(x+9)= 3(2x - 7)

ElenaW [278]

Answer: X= 57/2

Hope this helps!

3 0

3 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

4x-3=2x+7 algebraic form ​

4x-3=2x+7 algebraic form