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Verizon [17]
3 years ago
9

Students rented buses and vans to go to a hockey game. There were 8 vans and 12 buses for 360 students.

Mathematics
1 answer:
scoundrel [369]3 years ago
8 0

Answer:

A: 8v+12b=360

V= van

B= Bus

B: Each van=9

Need to ride in vans:72

Step-by-step explanation:

12 buses 24 in each bus that equals 288. 360-288=72

and if you need to know how many are in each van you divide 72 by the number of vans that would be 9 per van.

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Please help. i not smart
____ [38]

Answer:

It is hard to see the numbers.

But I am pretty sure it is 6x+6.

Step-by-step explanation:

Hope this helps! :D

4 0
2 years ago
Find an equivalent fractions for each given fraction 60/12
evablogger [386]

Very easy to do this.

60/12 is your fraction.

Multiply numerator and denominator by the same number.

60*2/12*2 = 120/24

120/24 = 5

You can also simplify.

60/12 = 5

12/12 = 1

5/1 = 5

60/12 = 5

8 0
3 years ago
Read 2 more answers
Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real
hoa [83]

Answer: 0=2x^2-7x-9\\0=4x^2-3x-1\\0=x^2-2x-8


Step-by-step explanation:

We know that the standard quadratic equation is  ax^2+bx+c=0

Let's compare all the given equation to it and , find discriminant.

1. a=2, b= -7, c=-9

b^2-4ac=(-7)^2-4(2)(-9)=49+72>0

So it has 2 real number solutions.

2. a=1, b=-4, c=4

b^2-4ac=(-4)^2-4(1)(4)=16-16=0

So it has only 1 real number solution.

3. a=4, b=-3, c=-1

b^2-4ac=(-3)^2-4(4)(-1)=9+16=25>0

So it has 2 real number solutions.

4. a=1, b=-2, c=-8

b^2-4ac=(-2)^2-4(1)(-8)=4+32=36>0

So it has 2 real number solutions.

5. a=3, b=5, c=3

b^2-4ac=(5)^2-4(3)(3)=25-36=-9

Thus it does not has real solutions.



5 0
3 years ago
Read 2 more answers
0.13x0.11=????????????????
andre [41]

Answer:

0.0143 in decimal form

Step-by-step explanation:

3 0
3 years ago
3 points lie on an xy plane, with each point representing where a particular person lives. Amy lives at point (2,3); Brian lives
geniusboy [140]
The equidistant point for all three, will be the "centroid" of their triangular locations
thus \bf \textit{Centroid of a Triangle}
\begin{array}{llll}
Amy
\begin{array}{llll}
(2,3)\\
x_1,y_1
\end{array}\quad Brian
\begin{array}{llll}
(0,7)\\
x_2,y_2
\end{array}\quad Claire
\begin{array}{llll}
(-2,3)\\
x_3,y_3
\end{array}\\ \quad \\\\\\
\left(\cfrac{x_1+x_2+x_3}{3}\quad ,\cfrac{y_1+y_2+y_3}{3}\quad \right)
\end{array} 

6 0
3 years ago
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