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sladkih [1.3K]
3 years ago
5

Members of a softball team raised $1513.50 to go to a tournament. They rented a bus for $961.50 and budgeted $46 per player for

meals. Write and solve an equation which can be used to determine xx, the number of players the team can bring to the tournament.
Mathematics
1 answer:
aniked [119]3 years ago
6 0

Answer: 1513.50=961.50+46x

The number of players the team can bring to the tournament = 12.

Step-by-step explanation:

Let x= the number of players the team can bring to the tournament.

Total money raised = Rent of bus + (Cost per player) (Number of players)

1513.50=961.50+46x\\\\\Rightarrow\ 46x=1513.50-961.50\\\\\Rightarrow\ 46x =  552\\\\\Rightarrow x=\dfrac{552}{46}\\\\\Rightarrow x=12

Hence, the number of players the team can bring to the tournament = 12.

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aleksandrvk [35]

Answer:

x^2-49

Step-by-step explanation:

when u multiply the numbers in parenthesis u get 49-7x+7x-x^2 and when u simplify that u get x^2-49

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3 years ago
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Answer:

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3 years ago
Solve the system of equations. y= x + 17, y= 3x+1
Lemur [1.5K]

Answer:

x=8, y=25. As a point it's (8,25)

Step-by-step explanation:

here is the system:

y=x+17

y=3x+1

notice how both of the equations equal y. Therefore, we can substitute one expression as y in the other expression. (It'll equal y=y, which is a true statement)

so it'll be:

3x+1=x+17

subtract 1 from both sides

3x=x+16

subtract x from both sides

2x=16

divide by 2

x=8

now, substitute 8 as x into one of the equations and solve for y

let's take the first one for example

y=8+17

y=25

so the answer is x=8, y=25. As a point it's (8,25)

Hope this helps!

7 0
3 years ago
Given the matrices: 1 2 A= 1 -1 2 1 1 B= 3 4 Calculate AB: C11 C12 [2.1] х 1 2 3 4 C21 C22 C11 = C12 = -2 C22 - C215 DONE​
eduard

Answer:

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

Step-by-step explanation:

Given the matrices

A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}

B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}

Calculating AB:

\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}

Multiply the rows of the first matrix by the columns of the second matrix

                                   =\begin{pmatrix}1\cdot \:1+\left(-1\right)\cdot \:3&1\cdot \:2+\left(-1\right)\cdot \:4\\ 2\cdot \:1+1\cdot \:3&2\cdot \:2+1\cdot \:4\end{pmatrix}

                                   =\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}

Hence,

\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}

Therefore,

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

5 0
3 years ago
Read 2 more answers
Pls help, due in 4 minutes
Shtirlitz [24]

Answer:

3.8s

Step-by-step explanation:

Solve for h=0, you will get t=3.75

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3 years ago
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