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KiRa [710]
2 years ago
7

x^{2} -50}{x^{2} +7x+10}" alt="\frac{2x^{2} -50}{x^{2} +7x+10}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Vikentia [17]2 years ago
4 0

Step-by-step explanation:

\frac{2 {x}^{2}  - 50}{ {x}^{2}  + 7x + 10 }

\frac{2( {x}^{2} - 25) }{(x + 5)(x + 2}

\frac{2(x + 5)(x - 5)}{(x + 5)(x  +  2)}

\frac{2(x - 5)}{x + 2}

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Determine the slope of a
AysviL [449]

Answer:

<em>m</em> = 3/5

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Coordinates (x, y)
  • Slope Formula: \displaystyle m = \frac{y_2-y_1}{x_2-x_1}
  • Perpendicular lines are the negative reciprocal of the given line

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point G(-2, 3)

Point H(1, -2)

<u>Step 2: Find slope </u><em><u>m</u></em>

Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>

<em />

<em>Line GH</em>

  1. Substitute in points [Slope Formula]:                                                              \displaystyle m = \frac{-2-3}{1--2}
  2. [Fraction] Subtract:                                                                                           \displaystyle m = \frac{-5}{3}

<em>Perpendicular line</em>

  1. Negative:                                                                                                           \displaystyle m = \frac{5}{3}
  2. Reciprocate:                                                                                                      \displaystyle m = \frac{3}{5}
4 0
2 years ago
Is the quotient of two rational numbers always a rational number? explain.
lys-0071 [83]

Answer:

yes because rational numbers can be written as a fraction

Step-by-step explanation:

7 0
3 years ago
I have a few maths questions I need help with, please bare with me
Alexeev081 [22]

Answer:

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5 0
3 years ago
Plz halp. I just need #20, and please write out the equation so I know how to do it for future problems, thanks!
FrozenT [24]
The number you need is 16 all you have to do is work out the problem and try to understand it they are trying to confuse you don't need an equation.
Hope this help!
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6 0
3 years ago
The line 5x – 5y = 2 intersects the curve x2y – 5x + y + 2 = 0 at
inna [77]

Answer:

(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve at each point of intersection are;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28

Step-by-step explanation:

The equations of the lines are;

5·x - 5·y = 2......(1)

x²·y - 5·x + y + 2 = 0.......(2)

Making y the subject of equation (1) gives;

5·y = 5·x - 2

y = (5·x - 2)/5

Making y the subject of equation (2) gives;

y·(x² + 1) - 5·x + 2 = 0

y = (5·x - 2)/(x² + 1)

Therefore, at the point the two lines intersect their coordinates are equal thus we have;

y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)

Which gives;

\dfrac{5 \cdot x - 2}{5} = \dfrac{5 \cdot x - 2}{x^2 + 1}

Therefore, 5 = x² + 1

x² = 5 - 1 = 4

x = √4 = 2

Which is an indication that the x-coordinate is equal to 2

The y-coordinate is therefore;

y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5

The coordinates of the points of intersection = (2, 8/5}

Cross multiplying the following equation

Substituting the value for y in equation (2) with (5·x - 2)/5 gives;

\dfrac{5 \cdot x^3 - 2 \cdot x^2 - 20 \cdot x + 8}{5} = 0

Therefore;

5·x³ - 2·x² - 20·x + 8 = 0

(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8

Therefore, we have;

x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²

5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8

∴ c = 8/(-2) = -4

2·b + c = - 20

b = -16/2 = -8

Therefore;

(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)

(x - 2)×(5·x² + 8·x - 4) = 0

5·x² + 8·x - 4 = 0

x² + 8/5·x - 4/5  = 0

(x + 4/5)² - (4/5)² - 4/5 = 0

(x + 4/5)² = 36/25

x + 4/5 = ±6/5

x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2

Hence the three x-coordinates are

x = 2, x = - 2, and x = 2/5

The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0

The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve, \dfrac{\mathrm{d} y}{\mathrm{d} x}, is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)

Therefore, we have;

\dfrac{\mathrm{d} y}{\mathrm{d} x}= \dfrac{\mathrm{d} \left (\dfrac{5 \cdot x - 2}{x^2 + 1}  \right )}{\mathrm{d} x} = \dfrac{5\cdot \left ( x^{2} +1\right )-\left ( 5\cdot x-2 \right )\cdot 2\cdot x}{\left (x^2 + 1 ^{2} \right )}.......(3)

Which gives by plugging in the value of x in the slope equation;

At x = -2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.92

At x = 2/5, \dfrac{\mathrm{d} y}{\mathrm{d} x} = 4.3

At x = 2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.28

Therefore;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28.

7 0
3 years ago
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