Answer:
$0 < p ≤ $25
Step-by-step explanation:
We know that coach Rivas can spend up to $750 on 30 swimsuits.
This means that the maximum cost that the coach can afford to pay is $750, then if the cost for the 30 swimsuits is C, we have the inequality:
C ≤ $750
Now, if each swimsuit costs p, then 30 of them costs 30 times p, then the cost of the swimsuits is:
C = 30*p
Then we have the inequality:
30*p ≤ $750.
To find the possible values of p, we just need to isolate p in one side of the inequality.
So we can divide both sides by 30 to get:
(30*p)/30 ≤ $750/30
p ≤ $25
And we also should add the restriction:
$0 < p ≤ $25
Because a swimsuit can not cost 0 dollars or less than that.
Then the inequality that represents the possible values of p is:
$0 < p ≤ $25
Answer:
the points are (35,30) you may need to search a graph online i used desmos
Step-by-step explanation:
1. x= Jacksons Cups y= Lucius Cups2. x - y= 5 6x + 3y = 300Substitution: 6 (y+5) + 3y = 300 6y + 30 + 3y = 300 9y = 270 dived both by 9 y=30 Sub y for other equation x - 30 = 5 add 30 to both sides x = 35 Answer: (35, 30) Graphing: x- y = 56x+ 3y = 300solve both for y y = x-5 6x + 3y = 300minus 6x from both sides the points are in y = mx + b y= 1x -5 y= -2x + 100 I will also leave the ss of the graph in the comments if you cannot see it My labels for the x-axis is Jacksons cups and y Is luscious cups. Elimination:x - y = 5 6x + 3y = 300 First I manipulated the equations by the following - 6 (x - y = 5 ) 1(6x + 3y = 300 ) -6x + 6y = -30 6x + 3y = 300 The 6 x's cancel and add the y's and real numbers together 9y = 270 dived both by 9 y= 30 Sub y for other equation x - 30 = 6 add 30 to both sides x= 30 The points are (35, 30) The solution is (35,30)They represent how many cups they sold. 35 is Jackson cups and 30 is Lucious's cups
(16/18)
(24/27)
You just times 8 and 9 by the same number, like:
(8•2/9•2)=(16/18)
(8•3/9•3)=(24/27)
You could do any number:
(8•100/9•100)=(800/900)
Answer:
-5
Step-by-step explanation:
Let n be the unknown number
difference is subtraction
3n -4 = -19
Add 4 to each side
3n-4+4 = -19+4
3n = -15
Divide by 3
3n/3 = -15/3
n = -5
I'm assuming you're talking about the indefinite integral

and that your question is whether the substitution

would work. Well, let's check it out:



which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)
What if we tried

next? Then

, giving

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.
So how to deal with this integral? The answer lies in what's called the "error function" defined as

By the fundamental theorem of calculus, taking the derivative of both sides yields

and so the antiderivative would be

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.