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netineya [11]
3 years ago
7

In how many months will the total cost of both options be the same? What will that cost be?

Mathematics
1 answer:
gregori [183]3 years ago
7 0
If you make yourself a monthly expense chart:
                                                       Months
                    <u>  1       2            3           4        5        6       7      8        9        10</u>
Opt 1         -80      -30        -30        -30      -30     -30    -30    -30   -30      -30
Opt 2         -40     -40        -40        -40       -40      -40    -40    -40    -40   -40

By month 5 the total amount spent for Opt 1 is $200 and Opt 2 is $200

For part two add up the values for each all the way up to month 9.
Option 1:   80 + 30(8) =  80 + 240 = 320 spent
Option 2:   40(9)  = 360  spent

So option 1 is cheaper by month 9

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3 years ago
What is the answer to this 14-2y=​
yaroslaw [1]

Answer:

y=7

Step-by-step explanation:

Assuming that the equation is 14-2y=0

We need to isolate 'y' variable to obtain its value, to do that, lets start by subtracting -14 to both sides of equality sign,

14-2y=0 is now this: 14-14-2y=0-14, which is equals to -2y=-14

Now, lets continue by dividing by -2 both sides of equality sign.

We have this

-2y/-2=-14/-2

y=7

Thus, we've obtained the value for 'y' variable

4 0
3 years ago
A 12-pound bag weighs more than a 200-ounce bag. True or false
Anastasy [175]

Answer:

false

Step-by-step explanation:

200oz= 12.5 pounds

7 0
3 years ago
Read 2 more answers
Find the quotient
Sindrei [870]

Answer:

1) 2.3

2) 4.088

3) .416

4) 12.52

5) 1012.5

Step-by-step explanation:

8 0
3 years ago
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the polynomial ky^3+3y^2-3 and 2y^3-5y+k when divided by (y-5) leave the same remainder in each case. Find the value of k.
marshall27 [118]

Answer:

k=\frac{153}{124}

Step-by-step explanation:

According to the Remainder Theorem; when P(y) is divided by y-a, the remainder is p(a).

The first polynomial is :

p(y)=ky^3+3y^2-3

When p(y) is divided by y-5, the remainder is

p(5)=k(5)^3+3(5)^2-3

p(5)=125k+75-3

p(5)=125k+72

When the second polynomial:

m(y)= 2y^3-5y+k is divided by y-5.

The remainder is;

m(5)= 2(5)^3-5(5)+k

m(5)= 225+k

The two remainders are equal;

\implies 125k+72=225+k

\implies 125k-k=225-72

\implies 124k=153

k=\frac{153}{124}

3 0
3 years ago
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