Answer:
Step-by-step explanation:
If we take out the extra $3, we can group the bills into one each of $5 and $1, for a value of $6. There will be 7 such groups in the remaining $42.
That means there are 7 bills of the $5 denomination, and 3 more than that (10 bills) of the $1 denomination.
There are 7 $5 bills and 10 $1 bills.
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If you want to write an equation, it is usually best to let a variable stand for the most-valuable contributor. Here, we can let x represent then number of $5 bills. Then the value of the cash box is ...
5x +(x+3) = 45
6x = 42 . . . . . . . . subtract 3, collect terms
x = 7 . . . . . . . . . . . there are 7 $5 bills
x+3 = 10 . . . . . . . . there are 10 $1 bills
You may notice that this working parallels the verbal description above. (After we subtract $3, x is the number of $6 groups.)
Answer:
48%
Step-by-step explanation:
48 out of 100 = 48%
:D
Answer:
Step-by-step explanation:
y=(x+5)2−1
Use the vertex form,
y=a(x−h)2+k, to determine the values of a, h, and .a=1h=−5k=−1Find the vertex(h,k(−5,−1)
Confused what the question is. Are you looking for the product or the zeroes?
If you are looking for the product, then:
Use foil to get: sec²(1) - sec²(-csc²) -1(1) -1(-csc²)
= sec² + sec²csc² - 1 + csc²
= sec²csc² + sec² + csc² - 1
= sec²csc² + 1 - 1 (NOTE: sec² + csc² = 1 is an identity)
= sec²csc²
Answer: sec²csc²
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If you are looking for the zeroes, then:
Using the zero product property, set each factor equal to zero and solve.
<u>First factor:</u>
sec²Θ - 1 = 0
sec²Θ = 1
secΘ = 1, -1
remember that secΘ is 
= 1 = -1
cross multiply to get:
cosΘ = 1 cosΘ = -1
use the unit circle (or a calculator) to find that Θ = 0 and π
<u>Second factor:</u>
1 - csc²Θ = 0
1 = csc²Θ
1, -1 = cscΘ
remember that cscΘ is 
= 1 = -1
cross multiply to get:
sinΘ = 1 sinΘ = -1
use the unit circle (or a calculator) to find that Θ = 
and 
Answer: 0, π, ,
Its 25.2 but what do u mean by that
