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ss7ja [257]
3 years ago
8

Can somebody help me out please?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer:

all work is shown and pictured

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Help with this question
Sholpan [36]

With some simplification,

cos⁻¹(x) - 5 cos⁻¹(x) - π/3 = 2π/3

becomes

-4 cos⁻¹(x) = π

or

cos⁻¹(x) = -π/4

However, the inverse cosine function has a range of 0 ≤ cos⁻¹(x) ≤ π, so there are no solutions to this equation.

4 0
3 years ago
Divide.
Elenna [48]

Answer:

51715x/19x

Step-by-step explanation:

6 0
3 years ago
200 meters
Katyanochek1 [597]

Answer:

520 m

Step-by-step explanation:

Perimeter is the distance around a shape. To find the perimeter, you add up all the sides.

200 + 60 + 60 + 200 = 520 m

7 0
3 years ago
Solve the equation x2 = 80.
mixer [17]

Answer:

x^2 = 80

x = √80

x = 8.9

........

5 0
3 years ago
What is a cubic polynomial function with zeros 3,3 and -3? show your work.​
Brums [2.3K]

Answer:

Cubic polynomial function with zeros 3,3 and -3 is \mathbf{x^3-3x^2-9x+27=0}

Step-by-step explanation:

We need to find a cubic polynomial function with zeros 3,3 and -3.

If zeros of polynomial are: 3,3,and -3

we can write:

x=3, x=3, x=-3

Or

We can write:

x-3=0, x-3=0, x+3=0

Now, we can write them as:

(x-3)(x-3)(x+3)=0

Multiplying the terms, we can find the polynomial:

(x-3)(x-3)(x+3)=0\\(x(x-3)-3(x-3))(x+3)=0\\(x^2-3x-3x+9)(x+3)=0\\(x^2-6x+9)(x+3)=0\\x(x^2-6x+9)+3(x^2-6x+9)=0\\x^3-6x^2+9x+3x^2-18x+27=0\\x^3-6x^2+3x^2+9x-18x+27=0\\x^3-3x^2-9x+27=0

So, cubic polynomial function with zeros 3,3 and -3 is \mathbf{x^3-3x^2-9x+27=0}

4 0
3 years ago
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