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Alexxx [7]
3 years ago
14

Which equation is a point slope form equation for line AB ?

Mathematics
1 answer:
Jobisdone [24]3 years ago
3 0

For this case we have that by definition, the equation of a line of the point-slope form is given by:

y-y_ {0} = m (x-x_ {0})

Where:

m: It's the slope

(x_ {0}, y_ {0}):It is a point through which the line passes

To find the slope, we need two points through which the line passes, observing the image we have:

(x_ {1}, y_ {1}): (1,6)\\(x_ {2}, y_ {2}): (5, -2)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-2-6} {5-1} = \frac {-8} {4} = -2

Thus, the equation is of the form:

y-y_ {0} = - 2 (x-x_ {0})

We choose a point:

(x_{0}, y_ {0}) :( 5, -2)

Finally, the equation is:

y - (- 2) = - 2 (x-5)\\y + 2 = -2 (x-5)

Answer:

y + 2 = -2 (x-5)

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C

Step-by-step explanation:

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Nadya [2.5K]
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8 0
2 years ago
What is the product of the two integer values for $x$ for which $|x^2 - 16|$ is a prime number?
Nana76 [90]

Answer:

-9

Step-by-step explanation:

x² -16 = (x -4)(x +4)

This magnitude of this product can only be prime if one of the factors is ±1. Any other integer value of x will produce a composite number (or zero).

... For x-4 = ±1, x = 3 or 5

... For x+4 = ±1, x = -3 or -5

The values of x that are ±3 both give |x²-16| = 7, a prime.

The values of x that are ±5 both give |x²-16| = 9, not a prime.

The two values of x that are of interest are x=-3 and x=3. Their product is ...

... (-3)·(3) = -9

8 0
3 years ago
Solve the following equation:
Rama09 [41]

Complete the square.

z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0

\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4

Use de Moivre's theorem to compute the square roots of the right side.

w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)

\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2

Now, taking square roots on both sides, we have

z^2 + \dfrac12 = \pm w^{1/2}

z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2

Use de Moivre's theorem again to take square roots on both sides.

w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)

\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}

w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)

\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}

3 0
1 year ago
When a = 6 and b = 22, c = 33. If c varies directly with b and inversely with a, which equation models the situation?
deff fn [24]

Answer:

c= 9b/a

Step-by-step explanation:

The statement "y varies directly as x," means that when x increases, y increases by the same factor. The statement "y varies inversely as x means that when x increases, ydecreases by the same factor.

Then, we can say that:

c ∝ b, thus c= kb and c = k/a.

Using both equations we have that:

c= kb/a

Given that a=6, b=22 and c=33 we have that:

33 = 22k/6

Solving for k:

k=9.

Then the equation that models the situation is:

c= 9b/a

4 0
3 years ago
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