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viva [34]
3 years ago
8

Find all of the zeros algebraically

Mathematics
1 answer:
erastova [34]3 years ago
7 0

Answer:

-4, 9, -1

Step-by-step explanation:

x³ - 4x² - 41x - 36

= x²(x + 4) - 8x² - 41x - 36

= x²(x + 4) - 8x(x + 4) - 9x - 36

= x²(x + 4) - 8x(x + 4) - 9(x + 4)

= (x + 4)(x² - 8x - 9)

= (x + 4) (x - 9)(x + 1)

so the zeros of f(x) are: -4, 9, -1

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Can anyone help me with this problem ? I would appreciate it a lot !
Scorpion4ik [409]

Answer:

D - 192

Step-by-step explanation:

So the solve this, lets use the formula for area of a triangle:

b*h / 2 = A

b is the base, h is the height.

In this case, our base is 32, and our height is unknown.

To find it we must use pythagreons theorm:

a^2+b^2=C^2

a is half of the base in this case, since tecnically this iscoceles traingle(2 same sides) cut in half is 2 right triangles.

32/2 = 16

b is the unknown, and C is the hypotenuse(longest length) which is 20:

16^2+b^2=20^2

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b^2=144

b=12

So we know b, whihc is our height, is 12.

Plugging this in:

32*12 / 2 = A

384 / 2 = A

192 = A

So the area of the triangle is 192.

Hope this helps!

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3 years ago
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ruslelena [56]

Answer:

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slavikrds [6]

Answer:

x = 1.

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y = k(x + 3)

When x = 3 y = 12 so

12 = k(3 + 3)

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6 0
3 years ago
Hi please help me fast thank You
son4ous [18]

Answer:

k=1/4

Step-by-step explanation:

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Very easy

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