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avanturin [10]
3 years ago
10

PLEASE HELP ASAP. GIVING BRAINLIEST

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
3 0

The equation is y = x + 35.275.

Step-by-step explanation:

Step 1; Gizmo starts at $425 and the monthly rate of return is 8.3%. So every month he gets back 8.3% of the $425 he starts with. To calculate how much 8.3% of $425 is we make 8.3% a fraction by dividing it by 100 and multiplying it with the amount.

8.3% of $425 = \frac{8.3}{100} × $425 = 0.083 × $425 = $35.275.

So every month $35.275 is added to the balance of the account.

Step 2; Assume y is the amount of money in the account at the end of the month while x is the amount of money at the end of the previous month's end. Using x and y we can form the following equation,

y = x + 35.275.

For the first month,       y = x + 35.725 = 425 + 35.725 = $460.725.

For the second month, y = x + 35.725 = 460.725 + 35.725 = $496.45.

For the third month,     y = x + 35.725 = 496.45 + 35.725 = $532.175.

For the fourth month,   y = x + 35.725 = 532.175 + 35.725 = $567.90.

For the fifth month,      y = x + 35.725 = 567.90 + 35.725 = $603.625‬.

For the sixth month,     y = x + 35.725 = 603.625‬ + 35.725 = $639.35.‬

For the seventh month, y = x + 35.725 = 639.35‬ + 35.725 = $675.075‬.

For the eighth month,    y = x + 35.725 = 675.075‬ + 35.725 = $710.80.

For the ninth month,     y = x + 35.725 = 710.80 + 35.725 = $746.525‬.

For the tenth month,     y = x + 35.725 = 746.525‬ + 35.725 = $782.25.

For the eleventh month,y = x + 35.725 = 782.25 + 35.725 = $817.975‬.

For the twelfth month,    y = x + 35.725 = 817.975‬ + 35.725 = $853.70

Gizmo then adds $250 to the $853.70 at the beginning of the second year‬.

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Use the rational zeroes theorem to state all the possible zeroes of the following polynomial:
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Answer:

All the possible zeroes of the polynomial: f(x) = 3x^{6} + 4x^{3} - 2x^{2} +4 are  ±1 , ±2 ,  ±4 ,  ±\frac{1}{3} , ±\frac{2}{3}  , ±\frac{4}{3} by using rational zeroes theorem.

Step-by-step explanation:

Rational zeroes theorem gives the possible roots of polynomial f(x) by taking ratio of p and q where p is a factor of constant term and q is a factor of the leading coefficient.

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Find all factors (p) of the constant term.

Here we are looking for the factors of 4, which are:

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Now find all factors (q) of the coefficient of the leading term

we are looking for the factors of 3, which are:

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List all possible combinations of ± \frac{p}{q}  as the possible zeros of the polynomial.

Thus, we have ±1 , ±2 ,  ±4 ,  ±\frac{1}{3} , ±\frac{2}{3}  , ±\frac{4}{3} as the possible zeros of the polynomial

Simplify the list to remove and repeated elements.

All the possible zeroes of the polynomial: f(x) = 3x^{6} + 4x^{3} - 2x^{2} +4 are  ±1 , ±2 ,  ±4 ,  ±\frac{1}{3} , ±\frac{2}{3}  , ±\frac{4}{3}

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I'm 87% sure the answer is B

6 0
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What is m∠DFE?<br><br> A. 78<br><br> B. 42<br><br> C. 19<br><br> D. 119
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---------------------------------------

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