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3 years ago

5.

Mathematics

1 answer:

kari74 [83]3 years ago

3 0

The length of the vector <-4,-3> is

$\sqrt{(-4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$

So, if we divide the components by 5, we'll get a vector with unitary length:

$v=\left(-\dfrac{4}{5}, -\dfrac{3}{5}\right) \implies |v| = \sqrt{\left(-\dfrac{4}{5}\right)^2+\left(-\dfrac{3}{5}\right)^2}=\sqrt{\dfrac{16}{25}+\dfrac{9}{25}}=\sqrt{\dfrac{25}{25}}=1$

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The remainder when p\left(x\right)=x^3-2x^2+8x+kp ( x ) = x 3 − 2 x 2 + 8 x + k by (x-2) is 19. What is the remainder when p(x)

Dafna11 [192]

Answer:

Remainder when p(x) is divided by (x+2) is -29

Step-by-step explanation:

p(x) = $x^{3} - 2x^{2} + 8x + k$

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p(x - 2 = 0) gives the remainder when p(x) is divided by (x-2)

x - 2 = 0

x = 2

p(x-2=0) = p(2) = $2^{3} - 2(2^{2}) + 8(2) + k$ = 19

8 - 8 + 16 + k = 19

k = 3

p(x) = $x^{3} - 2x^{2} + 8x + 3$

p(x + 2 = 0) gives the remainder when p(x) is divided by (x+2)

x + 2 = 0

x = -2

p(x+2=0) = p(-2) = $(-2)^{3} - 2((-2)^{2}) + 8(-2) + 3$

p(-2) = - 8 - 8 - 16 + 3 = -29

Remainder when p(x) is divided by (x+2) is -29

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3 years ago

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Which single transformation can be applied to the graph of y = 72 StartRoot x EndRoot to produce the graph of y = 9 StartRoot x

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Answer:

vertical....

Step-by-step explanation:

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The first AMC 8 was given in 1985 and it has been given annually since that time. Samantha turned 12 years old the year that she

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I would start like this: which year was the seventh AMC 8??

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The second: 1986
3: 1987
4: 1988
5: 1989
6: 1990
7: 1991

so seventh AMC 8 took place in 1991.
she was then 12, so we can subtract this age from 1991:

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The probability is 0.9671.

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z = (x - μ)/(σ/√n)
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For problems 29 - 31 the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c

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Answer:

29) discriminant is positive

30) discriminant is 0

31) discriminant is negative

Step-by-step explanation:

the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c = 0 is positive, negative, or zero.

In the graph of question number 29 we can see that the graph intersects the x axis at two points

so the equation has 2 solutions.

When the equation has two solution then the discriminant is positive

In the graph of question number 30 we can see that the graph intersects the x axis at only one point

so the equation has only 1 solution.

When the equation has only one solution then the discriminant is equal to 0

In the graph of question number 30 we can see that the graph does not intersects the x axis

so the equation has 2 imaginary solutions.

When the equation has two imaginary solutions then the discriminant is negative

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