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babymother [125]
3 years ago
8

Liam had $250. Then, he and his classmates bought a present for their teacher, evenly splitting the $120 among the e of them. Ho

w much money does Liam have left? Write your answer as an expression.
Mathematics
2 answers:
Gemiola [76]3 years ago
7 0
Liam Splits $40 With The 3 Of Them Out Of $120 And Liam Has 130 Because 250-120=130
garri49 [273]3 years ago
7 0

Answer:

250-\frac{120}{e}

Step-by-step explanation:

We have been given that Liam and his classmates bought a present for their teacher, evenly splitting the $120 among the 'e' of them.

This means that number of all students is 'e'. To find the cost of present for each student, we will divide 120 by e.

\text{Cost of present for each student}=\frac{120}{e}

Since Liam had $250, so we need to subtract \frac{120}{e} from 250 to get amount Liam has left:

\text{Amount Liam has left}=250-\frac{120}{e}

Therefore, our required expression would be 250-\frac{120}{e}.

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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
Evaluate -3x^3 - 4x for x=-1.
Ivenika [448]

Answer:

-x.(3x^2+4)

Step-by-step explanation:

Brain-list?

6 0
3 years ago
Read 2 more answers
Solve please. Correctly<br> 3(1 - 2a) = -3a - 18
True [87]

Answer:

a = 7

Step-by-step explanation:

3(1 - 2a) = -3a - 18

3 - 6a = - 3a - 18

Add 18 to each side:

3 + 18 - 6a = - 3a - 18 + 18

21 - 6a = - 3a

Add 6a to each side:

21 - 6a + 6a = - 3a + 6a

21 = 3a

3a = 21

Divide each side by 3:

3a ÷ 3 = 21 ÷ 3

a = 7

6 0
2 years ago
Solve the equation below and find the variation constant, k .Find y when x = 7 and z=5, if y varies jointly as x and 2, and y= 1
Vlada [557]
<h3>Answers:</h3>

Variation Constant: k = 7/6

y = 245/6 when x = 7 and z = 5

====================================

Explanation:

"y varies jointly as x and z" means that y = kxz

We have (x,y,z) = (4,14,3) as one triple, so,

y = kxz

14 = k*4*3

14 = 12k

12k = 14

k = 14/12

k = 7/6 is the variation constant

The equation goes from y = kxz to y = (7/6)xz

----------------

Plug in x = 7 and z = 5 to find y

y = (7/6)xz

y = (7/6)*7*5

y = (7/6)*(7/1)*(5/1)

y = (7*7*5)/(6*1*1)

y = 245/6

4 0
3 years ago
Henry and 28 classmates go to the roller skating rink . Each van can hold 11 students . If all of the vans are full except one h
Rudik [331]
You divide 29:11= 2 7/11
so you have 2 full vans and one with 7 students.
so the answer is 7
8 0
3 years ago
Read 2 more answers
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