<h3>
Answer:</h3>
- A) p = 5, one solution
- B) no solutions
- C) infinite solutions
<h3>
Step-by-step explanation:</h3>
A) Add 19-5p to each side of the equation:
... 10 = 2p
... 5 = p . . . . . divide by the coefficient of p
B) Subtract 5p from both sides of the equation:
... -9 = -19 . . . . . there is <em>no value of p</em> that will make this true. (No solution.)
C) Subtract 5p from both sides of the equation:
... -9 = -9 . . . . . this is true for <em>every value of p</em>. (Infinite solutions.)
What are you trying to do here?
Solve the graph, or make it appear as something else?
First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0
sec (x) (2sec (x) -2) = 0
Then we're going to separate the two to find the zeros of each because anything time 0 is zero.
sec(x) = 0
2sec (x) - 2 = 0
Now, let's simplify the second one as the first one is already.
Add 2 to both sides:
2sec (x) = 2
Divide by 3 on both sides:
sec (x) = 1
I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!
Let "a" and "b" represent the values of the first and second purchases, respectively.
0.40*(original price of "a") = $10
(original price of "a") = $10/0.40 = $25.00 . . . . divide by 0.40 and evaluate
a = (original price of "a") - $10 . . . . . . Julia paid the price after the discount
a = $25.00 -10.00 = $15.00
At the other store,
$29 = 0.58b
$29/0.58 = b = $50 . . . . . . . divide by the coefficient of b and evaluate
Then Julia's total spending is
a + b = $15.00 +50.00 = $65.00
Julia spent $65 in all at the two stores.
The cost of one ticket is $0.75
