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STALIN [3.7K]
3 years ago
7

3. The sum of two numbers is 17. The difference between the larger number and the

Mathematics
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

5

Step-by-step explanation:

To solve this problem, we need to set up a system of equations. Let x represent the larger number and y represent the smaller number.

x + y = 17

x - y = 7

We can solve this system by adding them together. The y cancels out because y + (-y) = 0. After that, we solve for x

2x = 24          

x = 12              (Divide both sides by 2)

Now that we know the value of x, we can substitute it into one of our original equations to solve for y.

12 + y = 17      

y = 5                (Subtract 12 from both sides)

So, the smaller number is 5.

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I will do a step by step explanation to help you.

First, subtract 16 from 31 so you can simplify what you need to find. We will call the unknown number X. So 5x = 15

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Divide 9x ^ 3 + 18x ^ 2 - 13x + 5 by 3x - 1 using division and write the division in the form P = DQ + R
lana66690 [7]

Answer:

\frac{9x^3\:+\:18x\:^2\:-\:13x\:+\:5}{3x-1}=9x^3+18x^2-13x+5

Step-by-step explanation:

DIVISION ALGORITHM: If p(x) and d(x)\neq 0  are polynomials, and the degree of d(x) is less than or equal to the degree of f(x),  then there exist unique polynomials q(x) and r(x), so that

                                               \frac{p(x)}{d(x)} =q(x)+\frac{r(x)}{d(x)}

and so that the degree of r(x)  is less than the degree of d(x).

To find \frac{9 x^{3} + 18 x^{2} - 13 x + 5}{3 x - 1} you must:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}9x^3+18x^2-13x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{9x^3}{3x}=3x^2

\mathrm{Quotient}=3x^2

\mathrm{Multiply\:}3x-1\mathrm{\:by\:}3x^2:\:9x^3-3x^2\\\mathrm{Subtract\:}9x^3-3x^2\mathrm{\:from\:}9x^3+18x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\

\mathrm{Remainder}=21x^2-13x+5

Therefore,

\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+\frac{21x^2-13x+5}{3x-1}

\mathrm{Divide}\:\frac{21x^2-13x+5}{3x-1}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}21x^2-13x+54\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{21x^2}{3x}=7x\\\\\mathrm{Quotient}=7x

\mathrm{Multiply\:}3x-1\mathrm{\:by\:}7x:\:21x^2-7x\\\mathrm{Subtract\:}21x^2-7x\mathrm{\:from\:}21x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-6x+5

Therefore,

\frac{21x^2-13x+5}{3x-1}=7x+\frac{-6x+5}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x+\frac{-6x+5}{3x-1}

\mathrm{Divide}\:\frac{-6x+5}{3x-1}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{-6x}{3x}=-2\\\\\mathrm{Quotient}=-2

\mathrm{Multiply\:}3x-1\mathrm{\:by\:}-2:\:-6x+2\\\mathrm{Subtract\:}-6x+2\mathrm{\:from\:}-6x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=3

Therefore,

\frac{-6x+5}{3x-1}=-2+\frac{3}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x-2+\frac{3}{3x-1}

6 0
4 years ago
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