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Makovka662 [10]
3 years ago
9

A limited-edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years, it is worth $23.

81. Which equation can be used to find the value, y, after x years? (Round money values to the nearest penny.)
y = 18(1.15)x
y = 18(0.15)x
y = 20.7(1.15)x
y = 20.7(0.15)x
Mathematics
1 answer:
erica [24]3 years ago
4 0
Y = 18(1.15)x is the answer
After one year
18×(1.15)=20.7
After two years
18×(1.15)^(2)=23.81
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Nate rented a power washer and a generator. On his first job, he used each piece of equipment for 6 hours which cost $90. On his
sp2606 [1]

Answer:

Step-by-step explanation:

p = power washer and g = generator

6p + 6g = 90 .....multiply by 4

4p + 8g = 100 ....multiply by -6

---------------------

24p + 24g = 360 (result of multiplying by 4)

-24p - 48g = -600 (result of multiplying by -6)

---------------------add

-24g = - 240

g = -240 / -24

g = 10 <=== the generator cost $ 10 per hr

6p + 6g = 90

6p + 6(10) = 90

6p + 60 = 90

6p = 90 - 60

6p = 30

p = 30/6

p = 5 <==== the power washer costs $ 5 per hr

check...

4p + 8g = 100

4(5) + 8(10) = 100

20 + 80 = 100

100 = 100 (correct)

3 0
3 years ago
24y - 22 = 4 ( 6y - 6 )
Pie
I don’t think their is a solution to this equation
because if you expand the second half it is= 24y-24 which would make the equation
- 24y-22=24y-24
and because the number next to the y is the same on both sides, no matter what y is if we subtract different numbers from each side we will never get the same value for each side of the =
5 0
3 years ago
Mrs. Evans gave her class a 100-page reading assignment. After 2 days, Kya and
Zarrin [17]

Answer:

so the answer is 20 because u know how she take it for days

4 0
2 years ago
Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81
Anton [14]
Using Lagrange multipliers, we have the Lagrangian

L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)

with partial derivatives (set equal to 0)

L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}
L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}
L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}
L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81

Substituting the first three equations into the fourth allows us to solve for \lambda:

x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}

For each possible value of \lambda, we get two corresponding critical points at (\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3).

At these points, respectively, we get a maximum value of f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3 and a minimum value of f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3.
5 0
3 years ago
Can someone please help
Sloan [31]

Answer:

41

Step-by-step explanation:

an = a1 + (n-1)d

a_n = the nᵗʰ term in the sequence = 10

a_1 = the first term in the sequence = 5

d = the common difference between terms = 4

a10 = 5 + (10-1)4

a10 = 5 + (9)4

a10 = 5 + 36

a10 = 41

7 0
2 years ago
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