<u>Answer:</u>
<h2>
52% were jelly beans!</h2>
<u>Explanation</u><u>:</u>
<em>Cross </em><em>multiply </em><em>the </em><em>following:</em>
<em></em>
<em>x </em><em>times </em><em>1</em><em>3</em><em>8</em><em> </em><em>=</em><em> </em><em>1</em><em>3</em><em>8</em><em>x</em>
<em>100 </em><em>times </em><em>7</em><em>1</em><em>.</em><em>7</em><em>6</em><em> </em><em>=</em><em> </em><em>7</em><em>,</em><em>1</em><em>7</em><em>6</em>
<em>Divide </em><em>both </em><em>sides </em><em>by </em><em>1</em><em>3</em><em>8</em><em>:</em>
<em></em>
<h3>
<em>x </em><em>=</em><em> </em><em>5</em><em>2</em><em>%</em></h3>
Answer:
1 x=-2.5 y = -5.5
2. x=5 y=1
Step-by-step explanation:
1) What is the solution of the given system?
5x-y=-7
3x-y=-2
Multiply the second equation by -1
-1*(3x-y)=-1(-2)
-3x +y = 2
Now add the first equation to the modified second equation
5x-y=-7
-3x +y = 2
------------------
2x = -5
Divide each side by 2
2x/2 = -5/2
x = -2.5
Now we need to find y
-3x+y =2
-3(-2.5) +y =2
7.5 +y =2
Subtract 7.5 from each side
7.5 -7.5 +y =2-7.5
y = -5.5
2) what is the solution of the given system?
5x+7y=32
8x+6y=46
Divide the second equation by 2
8x/2+6y/2=46/2
4x+3y =23
Multiply the first equation by 4
4 (5x+7y)=32*4
20x+28y = 128
Now multiply the modified 2nd equation by -5
-5(4x+3y )=-5(23
)
-20x -15y = -115
Lets add the new equations together to eliminate x
20x+28y = 128
-20x -15y = -115
---------------------
13y = 13
Divide each side by 13
13y/13 =13/13
y=1
Now substitute back in to find x
5x+7y=32
5x +7(1) =32
5x +7 =32
Subtract 7 from each side
5x+7-7 =32-7
5x =25
Divide by 5
5x/5 =25/5
x=5
Answer:
Yes, an arrow can be drawn from 10.3 so the relation is a function.
Step-by-step explanation:
Assuming the diagram on the left is the domain(the inputs) and the diagram on the right is the range(the outputs), yes, an arrow can be drawn from 10.3 and the relation will be a function.
The only time something isn't a function is if two different outputs had the same input. However, it's okay for two different inputs to have the same output.
In this problem, 10.3 is an input. If you drew an arrow from 10.3 to one of the values on the right, 10.3 would end up sharing an output with another input. This is allowed, and the relation would be classified as a function.
However, if you drew multiple arrows from 10.3 to different values on the right, then the relation would no longer be a function because 10.3, a single input, would have multiple outputs.
Be:
Number of hours: n
<span>The cost of renting a bike for the first hour is $7:
n=1→f(n)=f(1)=$7
</span>He is charged $2.50 for every additional hour of renting the bike:
f(n)=f(n-1)+2.50, for <span>n ≥ 2
</span>
f(1)=7; f(n)=f(n-1)+2.50, for <span>n ≥ 2 (sixth option)
</span>
f(n)=f(1)+2.50(n-1)
f(n)=7+2.50(n-1)
f(n)=7+2.50n-2.50
f(n)=2.50n+4.50 (fifth option)
Answers:
Fifth option: f(n)=2.50n+4.50, and
Sixth option: f(1)=7; f(n)=f(n-1)+2.50, for <span>n ≥ 2</span>