PLEASE HELP!!!!!!!! ASAP

Mathematics

2 answers:

Varvara68 [4.7K]2 years ago

8 0

Question 1: The areas are equal
Question 2: The area of the circle would be 28.26 cm 2

inessss [21]2 years ago

3 0

Answer: #1) 3. The semicircle has twice the area of the circle

#2) 3. 28.26 cm

Step-by-step explanation:

#1) The formula for area of circle is A=πr².

First, plugging in the radius for the circle, A=π(5)², we get an area of about 78.54 inches².

Plugging in the radius for the semicircle, A=π(10)², which is equal to 314.159.... However, since it is a SEMI-circle, we will divide this value by 2 to give us an area of about 157.08 inches².

Comparing the two areas, we see the semicircle has twice the area of the circle as 78.54 x 2 = 157.08.

#2) Again use the area of circle formula, A=πr². The radius is half the diameter, so half of 6cm=3cm, meaning r=3. plug it in.

A= π(3)²

A= 28.27 cm²

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Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the func

mote1985 [20]

Solution:

The given Polynomial is :

$f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$

By Rational Root theorem the of Zeroes of the Polynopmial are:

$\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4$

But , $f(\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4)\neq 0$

So, no root of this polynomial is real.

Therefore, All the four roots of Polynomial are imaginary.

So, we can't say whether the number k=2, is an upper or lower bound of the polynomial $f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$ .