The product of this problem

1-Step Equations <br> 11 1/5+f=-2 1/2<br> (please help i’ve been trying to solve this for ages)

astra-53 [7]

Find the common denominator. It’s 10
22 2/10+f=-10 5/10
-22 2/10 -22 2/10
f = -32 7/10

3 0

3 years ago

Read 2 more answers

Jaden can read 18 pages in 1/3 of an hour. How many pages can he read in 2 hours

algol [13]

Answer is 108 pages. 1/3 of an hr is 20 min. 20*3=60 which is an hr. 3+3=6. So then you get 6*18=108.

8 0

3 years ago

Of the students at I.S. 59, 230 signed up for the school picnic. That was 50% of all students in the school. How many students a

Nastasia [14]

Answer: 460 students

Step-by-step explanation:

Based on the information given in the question, let the total number of students that attend I.S. 59 be represented by x.

Since we've been given the information that 230 signed up for the school picnic which was 50% of all students in the school. Then, the students in the school will be:

= 50% of x = 230

0.5 × x = 230

0.5x = 230

x = 230/0.5

x = 460

Therefore, 460 students attend the school.

4 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

Possible values for the area A of the rectangle shown are 12

KatRina [158]

Answer:

idc its 12

Step-by-step explanation:

5 0

3 years ago