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mylen [45]
3 years ago
5

Harry has $110 in his savings account. He decides to save $15 every month to buy a bike. An equation for this relation is C = 11

0 + 15m, where C represents the amount of money, in dollars, in his account and M represents the number of months. How many months does it take Harry to have $185 in his account?
Mathematics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

5 MONTHSSSSSS only 5

Step-by-step explanation:

K here we go

y=110+15x

185=110+15x

75=15x

x=5

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How many distinct positive integer-valued solutions exist to the equation (x2 - 7x + 11)(x2 - 13x + 42) = 1
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Answer:

The given equation has TWO positive integer valued solutions, {6, 7}

Step-by-step explanation:

Here we are given two trinomial factors, each of which needs to be set equal to zero and in each case the resulting quadratic equation solved.

(x^2 - 7x + 11) has the coefficients {1, -7, 11}, and so the discriminant of this quadratic is b^2 - 4(a)(c), or 49 - 4(11), or 5.

Because this discriminant is positive, we know immediately that this quadratic has two real, unequal roots involving √5 (NO integer roots).

Next we focus on (x^2 - 13x + 42).  The discriminant is b^2 - 4(a)(c), or

169 - 168, or 1.  Again we see that there are two real, unequal roots:

      +13 ± √1                  +13 ± 1

x = ---------------  or  x = -------------

             2                            2

OR x = 14/2 = 7 (integer) or x = 12/2 = 6 (integer).

The given equation has TWO positive integer valued solutions, {6, 7}

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3 years ago
When x = 3 and y = 5, by how much does the value of 3x^2 – 2y exceed the value of 2x^2 – 3y ?
JulsSmile [24]
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Solve for all complex roots (2x^10)+(3x^9)-(5x^8)-(9x^7)-(x^6)+(3x^5)+(7x^4)+(9x^3)-(x^2)-(6x)-2
fomenos

Solve for x by completing the square:

2 x^10 + 3 x^9 - 5 x^8 - 9 x^7 - x^6 + 3 x^5 + 7 x^4 + 9 x^3 - x^2 - 6 x - 2 = 0

The left hand side factors into a product with five terms:

(x - 1)^2 (x + 1)^3 (2 x + 1) (x^2 - 2) (x^2 + 1) = 0

Split into five equations:

(x - 1)^2 = 0 or (x + 1)^3 = 0 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Take the square root of both sides:

x - 1 = 0 or (x + 1)^3 = 0 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Add 1 to both sides:

x = 1 or (x + 1)^3 = 0 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Taking cube roots gives 0 times the third roots of unity:

x = 1 or x + 1 = 0 or x + 1 = 0 or x + 1 = 0 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x + 1 = 0 or x + 1 = 0 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x = -1 or x + 1 = 0 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x = -1 or x = -1 or 2 x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x = -1 or x = -1 or 2 x = -1 or x^2 - 2 = 0 or x^2 + 1 = 0

Divide both sides by 2:

x = 1 or x = -1 or x = -1 or x = -1 or x = -1/2 or x^2 - 2 = 0 or x^2 + 1 = 0

Add 2 to both sides:

x = 1 or x = -1 or x = -1 or x = -1 or x = -1/2 or x^2 = 2 or x^2 + 1 = 0

Take the square root of both sides:

x = 1 or x = -1 or x = -1 or x = -1 or x = -1/2 or x = sqrt(2) or x = -sqrt(2) or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x = -1 or x = -1 or x = -1/2 or x = sqrt(2) or x = -sqrt(2) or x^2 = -1

Take the square root of both sides:

x = 1 or x = -1 or x = -1 or x = -1 or x = -1/2 or x = sqrt(2) or x = -sqrt(2) or x = i or x = -i

There are {2} duplicate solutions:

Answer: x = -1 or x = 1 or x = -1/2 or x = -i or x = i or x = sqrt(2) or x = -sqrt(2)

3 0
3 years ago
Read 2 more answers
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