All Answers 358291 Answers

Like so:

Like so:

Like so:

This should help;

Consider this labelled diagram

It's given that \(OC=32 \)  and \(CB=38\)

⇒\(OB = OC+CB\)

           \(=32+38\)

           \(=70 \)

Radius of semi-circle (3) is \(70\). 

And, CB is diameter of semi-circle (2).

So, radius of semi-circle is \(38/2=19\). 

From figure, we can infer that

Ar. (semi-circle 1) = Ar. (semi-circle 3) - Ar. (semi-circle 2) 

                               \(= {4900\pi \over 2} - {361\pi \over 2}\)

                               \(={4539\pi \over 2}\)

                              \(=7126.23\) sq. units 

Thus, the area of circle with centre K is \(7126.23\) sq. units. 

~Amy 

In the diagram, angle U = 30, arc XY is 170, and arc VW is 110. Find arc WY, in degrees.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

arcVX + arcWY = 360 - (arcXY + arcVW)

∠U = 1/2(arcWY - arcVX)

In the diagram, angle U = 30, arc XY is 170, and arc VW is 110. Find arc WY, in degrees.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

arcVX + arcWY = 360 - (170 + 110)

30 = 1/2(arcWY - arcVX)

When x = y, ABCD is a square with side length 2x, and the radius is x.

By the Cosine Law, a^2 - ab + b^2 = c^2 \cos C.  Plug in similar expressions, and the rest is easy.

1. 0

1.2/3

https://web2.0calc.com/questions/cphill-please-help_6

Cphill has already answered this one. :))

=^._.^=

BA = (4, 4) with slope of -3. 

y = -3x + 16

OB = 16

DC = points (4, 4) and (8, 0)

y = -x + 8

OD = 8

DOC = 8*8/2 = 32

BDE = 8*4/2 = 16

16+32 = 48 :)

=^._.^=

It's solved now. :))

=^._.^=

L = x

B = x

I = 5 + x

______

L = x 

B = x - 10

I = x - 4

______

x - 10 = 2/5(x - 4)

x - 10 = 2x/5 - 8/5

5x - 50 = 2x - 8

3x = 42

x = 14

=^._.^=

f(1) shows us that the sum of the coeficients = 32

If a polynomial is ax^2 + bx + c, x = 1 will just be a + b + c. 

=^._.^=

5*3=15

Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. What is the degree of $f(x) \cdot g(x)$?

sorry...

Suppose $f$ is a polynomial such that $f(0) = 47$, $f(1) = 32$, $f(2) = -13$, and $f(3)=16$. What is the sum of the coefficients of $f$?

Hold up Hold up, which one do you want me to exactly do?