Consider the circle above. What is the approximate area of the circle?

Mathematics

1 answer:

Olin [163]3 years ago

4 0

Answer:

The area of a circle is : π (Pi) times the Radius squared : A = π r 2 or, when you know the Diameter: A = (π /4) × D 2 or, when you know the Circumference: A = C 2 / 4 π

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Evaluate 5 - (4 + 2)2

tresset_1 [31]

Answer:

-7

Step-by-step explanation:

We can follow PEMDAS and see that first we do 4+2 which is 6. Our equation looks like 5 - (6)2. Since multiplication comes before subtraction we multiply 6 * 2 which is 12. Now we can fit that in 5 - 12 = -7

4 0

3 years ago

Read 2 more answers

Please help me and answer correctly!

LenKa [72]

A^-b = 1/a^b
a^b / a^c = a^b-c
a^c / b^c =(a/b)^c

3 0

3 years ago

N=4; 2i and 3i are zeros <br> f(-1)=50

Nonamiya [84]

Solution:- $\text{Let f(x) be any nth degree polynomial with n=4}\\$

$\text{Given that 2i and 3i are the zeroes of f(x)}$

$\text{so (x-2i) and (x-3i) are factors of f(x)}$

$\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}$

$\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)$