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3 years ago

Find the perimeter of this quarter circle with radius,

Mathematics

1 answer:

Inga [223]3 years ago

6 0

Answer:

Step-by-step explanation:

First find the circumference of the circle

C = 2 * pi*r

C = 2 * pi *2

C = 4pi

A quarter circle is 1/4 of a circle

1/4 C = 1/4 (4pi)

1/4C = pi

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What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 + 5x3 − x2 + 8x + 4?

ale4655 [162]

Answer:

Either: 1 neg, 3 pos, 0 imaginary; 1 neg, 1 pos, 2 imaginary

Step-by-step explanation:

Look for the positive possibilities first. Count the numbe of sign changes then subtract 2, if possible, as many times as you can.

There are 3 sign changes. So the possible positive roots are either 3 or 1.

Now look for the negative possibilities. Replace each x with a -x and then count the sign changes. Replacing with -x's gives you this polynomial:

$f(-x)=-3x^4-5x^3-x^2-8x+4$

There is only one sign change here, so the possible negative roots is 1. Start with the negative roots to find the possible combinations of positive, negative, and imaginary, since there is only 1.

- 1 1

+ 3 1

i 0 2

Since this is a 4th degree, the number of roots we have has to add up to equal 4.

8 0

3 years ago

Use vectors to find the interior angles of the triangle with the given vertices. (Enter your answers as a comma-separated list.

SOVA2 [1]

Answer:

23.92°, 78.503°, and 77.577°

Step-by-step explanation:

The coordinates of the vertices of the triangle are;

X(-3, -5), Y(2, 6), Z(6, 3)

The vectors are;

z = $\left \langle 2 - (-3), 6 - (-5) \right \rangle$ = $\left \langle 5, 11 \right \rangle$

y = $\left \langle 6 - (-3), 3 - (-5) \right \rangle$ = $\left \langle 9, 8 \right \rangle$

x = $\left \langle 2 - 6, 6 - 3 \right \rangle$ = $\left \langle -4, 3 \right \rangle$

$cos (\alpha ) = \dfrac{z \cdot y}{\left | z \right | \times \left | y \right |}$

Therefore, we get;

$cos (\alpha ) = \dfrac{5 \times 9 + 11 \times 8 }{\left | \sqrt{5^2 + 11^2} \right | \times \left | \sqrt{9^2 + 8^2} \right |} = \dfrac{133}{\sqrt{146} \times \sqrt{145} } \approx 0.91409$

α = arccos(0.91490) ≈ 23.92°

γ = The angle between -y and x

-y = $\left \langle -9, -8 \right \rangle$

We get;

$cos (\gamma) = \dfrac{-y \cdot x}{\left | -y \right | \times \left | x \right |}$

Therefore;

$cos (\gamma) = \dfrac{-9 \times -4 + -8 \times 3 }{\left | \sqrt{(-9)^2 + (-8)^2} \right | \times \left | \sqrt{(-4)^2 + 3^2} \right |} = \dfrac{12}{\sqrt{145} \times \sqrt{25} } \approx 0.199309$