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V125BC [204]
3 years ago
6

Mr. Jamison deposited $100 into a new savings account on January 1. On the first day of each month thereafter, he deposited thre

e times the amount he deposited in the previous month. On June 15 of the same year, the total amount Mr. Jamison has deposited is $
Mathematics
2 answers:
tester [92]3 years ago
7 0

Answer:

36,400

Step-by-step explanation:

Lera25 [3.4K]3 years ago
5 0

Answer:

$36 400

Step-by-step explanation:

Step 1

The first step is to figure out how much money is saved at the end of each month for the period from January 1 to June 15. The amount deposited at the end of each month is obtained by multiplying the amount from the previous month by 3.

The amount deposited in January is  \$100.

The amount deposited in February is 1\$00\times 3= \$300.

The amount deposited in March is  \$300\times 3= \$900.

The amount deposited in April is  \$900\times 3= \$2\700.

The amount deposited in May is  \$2\,700\times 3= \$8\,100.

The amount deposited in June is  \$8\,100\times 3= \$24\,300.


Step 2

The next step is to add up all the money that was deposited into the account. This calculation is shown below,

\$100+\$300+900+\$\$2\,700+\$8\,100+\$24\,000=\$36\,400


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Solution:

we have been asked to verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3

To verify that whether the given values are zeros or not we will substitute the values in the given Polynomial, if it will returns zero, it mean that value is Zero of the polynomial. But if it return any thing other than zeros it mean that value is not the zero of the polynomial.

Let f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\

f( \frac{1}{2} ) = 4 ( \frac{1}{2} )^3+20(\frac{1}{2})^2+2(\frac{1}{2})-3=\frac{7}{2}\\\\

\text{when x=}\frac{3}{4}\\\\

f( \frac{3}{4} ) = 4 ( \frac{3}{4} )^3+20(\frac{3}{4})^2+2(\frac{3}{4})-3=\frac{183}{16}\\

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Since sum of roots=\frac{-b}{a}= \frac{-20}{4}=-5\\

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