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3 years ago

HELP ASAP 25 points. Vector question

Mathematics

1 answer:

Nutka1998 [239]3 years ago

5 0

Answer:

4v - 7w = (101 , -36)

6u -8v = (-78 , 58)

2u +v - 4w = (40 , -4)

11u + 3w = (-88 , 89).

Step-by-step explanation:

u(-5, 7) ; v(6 , -2) ; w(-11,4)

4v - 7w = (4*6+[-7]*[-11] , 4*[-2] + [-7]*4)

=(24+77 , -6 - 28)

4v - 7w = (101 , -36)

6u - 8v = (6*[-5]+[-8]*6 , 6*7+[-8]*[-2] )

= (-30-48 , 42+16)

6u -8v = (-78 , 58)

2u + v - 4w = (-10+6+44 , 14 -2 -16)

2u +v - 4w = (40 , -4)

11u + 3w = (11*[-5]+3*[-11] , 7*11 +3*4)

= (-55-33 , 77+12)

11u + 3w = (-88 , 89)

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Answer:

x = 1, -2/7

Step-by-step explanation:

You could use the quadratic equation but this can be factored into

(7x + 2) (x - 1). You can verify that by multiplying it out.

Since (7x + 2) (x - 1) = 0, if either factor is 0 then the equation would be equal to 0, thus we get x = 1, -2/7

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2 years ago

Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

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4 years ago

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