Answer:
- <em>To solve these first swap x and y, solve for y and name it inverse function</em>
3. <u>y = -8x + 2</u>
- x = -8y + 2
- 8y = -x + 2
- y = -x/8 + 2/8
- y = -(18)x + 1/4
f⁻¹(x) = -(18)x + 1/4
-----------------------------------------
4.<u> y = (2/3)x - 5</u>
- x = (2/3)y - 5
- (2/3)y = x + 5
- y = (3/2)x + (3/2)5
- y = 1.5x + 7.5
f⁻¹(x) = 1.5x + 7.5
-----------------------------------------
5. <u>f(x) = 2x² - 6</u>
- x = 2y² - 6
- 2y² = x + 6
- y² = 1/2x + 3
- y =
f⁻¹(x) =
-----------------------------------------
6. <u>y = (x - 3)²</u>
- x = (y - 3)²
- = y - 3
- y = 3 +
f⁻¹(x) = 3 +
Factor the denominator:
(3x-1)(x+2)
every time x=-2 (because of (x+2), it would cause a warp in the graph, and not have an input. that is the vertical asymptote.
since you have factor 3x-1 on top and bottom, in which the zero is 1/3, there would be no solution at x=1/3. it would be a hole.
the answer here is C.
Hello,
I note (a,b,c) the result of a quarters, b dimes and c pennies:
2 solutions:
106=( 3, 3, 1)=( 1, 8, 1)
106=( 0, 0, 106) but : 100= 0*25+ 0*10+ 100
106=( 0, 1, 96) but : 100= 0*25+ 1*10+ 90
106=( 0, 2, 86) but : 100= 0*25+ 2*10+ 80
106=( 0, 3, 76) but : 100= 0*25+ 3*10+ 70
106=( 0, 4, 66) but : 100= 0*25+ 4*10+ 60
106=( 0, 5, 56) but : 100= 0*25+ 5*10+ 50
106=( 0, 6, 46) but : 100= 0*25+ 6*10+ 40
106=( 0, 7, 36) but : 100= 0*25+ 7*10+ 30
106=( 0, 8, 26) but : 100= 0*25+ 8*10+ 20
106=( 0, 9, 16) but : 100= 0*25+ 9*10+ 10
106=( 0, 10, 6) but : 100= 0*25+ 10*10+ 0
106=( 1, 0, 81) but : 100= 1*25+ 0*10+ 75
106=( 1, 1, 71) but : 100= 1*25+ 1*10+ 65
106=( 1, 2, 61) but : 100= 1*25+ 2*10+ 55
106=( 1, 3, 51) but : 100= 1*25+ 3*10+ 45
106=( 1, 4, 41) but : 100= 1*25+ 4*10+ 35
106=( 1, 5, 31) but : 100= 1*25+ 5*10+ 25
106=( 1, 6, 21) but : 100= 1*25+ 6*10+ 15
106=( 1, 7, 11) but : 100= 1*25+ 7*10+ 5
106=( 1, 8, 1) is good
106=( 2, 0, 56) but : 100= 2*25+ 0*10+ 50
106=( 2, 1, 46) but : 100= 2*25+ 1*10+ 40
106=( 2, 2, 36) but : 100= 2*25+ 2*10+ 30
106=( 2, 3, 26) but : 100= 2*25+ 3*10+ 20
106=( 2, 4, 16) but : 100= 2*25+ 4*10+ 10
106=( 2, 5, 6) but : 100= 2*25+ 5*10+ 0
106=( 3, 0, 31) but : 100= 3*25+ 0*10+ 25
106=( 3, 1, 21) but : 100= 3*25+ 1*10+ 15
106=( 3, 2, 11) but : 100= 3*25+ 2*10+ 5
106=( 3, 3, 1) is good
106=( 4, 0, 6) but : 100= 4*25+ 0*10+ 0
Answer:
a.
b. His teacher will receive more pencils (See explanation).
Step-by-step explanation:
a. The total number of pencils Louis brings to school is:
Then, in order to calculate the number of pencils Louis’s teacher will receive after he gives each of his 15 classmates an equal number of pencils, you need to solve the division show in the picture attached.
Notice that the remainder obtained is: .
<em>This means that Louis’s teacher will receive 4 pencils.</em>
b. If Louis decides instead to take an equal share of the pencils along with his classmates, his teacher will receive more pencils; because the amount of pencils each classmate will receive will be less. This means that the number of pencils leftover will increase, leaving more pencils for his teacher.