All categories

3 years ago

What is the area of point ?

Mathematics

1 answer:

Fed [463]3 years ago

7 0

Answer:

314 cm²

Step-by-step explanation:

It's not the point O, but the circle O, a point has no area.

The area of a circle is given by the formula A = π r²

We don't have r (radius), but we have the diameter, which is double the radius. So, r = d / 2 = 20 cm / 2 = 10 cm

Now, we can plug r directly into the equation:

A = π r² = π 10² = 100π = 314 cm²

You might be interested in

Is (4, 8) a solution of y > 5x + 4?

lana66690 [7]

This is not a solution. When plugged in,
8 > 20 + 4.
8 > 24, which is not true. :)

8 0

3 years ago

Which set of ordered pairs does NOT represent a function?

Reika [66]

The answer is b because it repeats “2” two times

4 0

3 years ago

Please help with this question:

strojnjashka [21]

Answer:

yes

Step-by-step explanation:

as long as the two shortest sides add up to more than the longest side then a triangle can be made with those side lengths

8 0

3 years ago

Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5

givi [52]

<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

$A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]$ $X=\left[\begin{array}{c}x\\y\end{array}\right]$

and $B= \left[\begin{array}{c}11\\5\end{array}\right]$

∴AX=B

$adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]$

$A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6$

∴ $A^{-1} =\frac{adj A}{|A|}$

So, $A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}$

$A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}$

$X =A^{-1}\times B$

⇒ $\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]$

⇒ $\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c} {6}\\ {11} \end{array}\right]}$

∴ x= 6 and y = 11

The required solution is x = 6 and y = 11

7 0

3 years ago

Given z=4+3i evaluate z

DerKrebs [107]

Answer:

Step-by-step explanation:

Assuming we want to evaluate |z|, given that, z=4+3i.

Then, by definition of modulus,

$|x + yi| = \sqrt{ {x}^{2} + {y}^{2} }$

$|4+ 3i| = \sqrt{ {4}^{2} + {3}^{2} }$

$|4+ 3i| = \sqrt{ 16+ 9 } \\ |4+ 3i| = \sqrt{25 } \\ |4+ 3i| = 5$

Therefore the modulus be of the given complex number is 5 units

5 0

3 years ago