Find the area of the semicircle.

Mathematics

1 answer:

Oliga [24]3 years ago

7 0

So since this is half a circle, we have to divide the area of a circle by 2. So that would give us 1/2 pi* r^2. We find the radius by dividing 10 by 2. So the radius is 5. Plugging that in for r in 1/2(pi)(r^2) we get 1/2(pi)(5^2) which is 25pi/2. Turning pi into 3.14 we have (25(3.14))/2 which is 78.5/2 which is approximately 39.25 units^2

You might be interested in

Convert 6 ft. 5 in.to inches

trapecia [35]

Your answer is 77 inches

8 0

3 years ago

Read 2 more answers

A jar contains four black buttons and five brown buttons. If five buttons are picked at random, what is the probability that at

lbvjy [14]

Answer:

189/572=33.042%

5 0

4 years ago

Which is the value of this expression when a = negative 2 and b = negative 3? (StartFraction 3 a Superscript negative 3 Baseline

natta225 [31]

Answer: The answer is D

729 over 64

4 0

4 years ago

Read 2 more answers

4. What is the percent of 88 is 33?

IRISSAK [1]

33 = 88x
x = 33/88
x = .375 = 37.5 percent

6 0

3 years ago

Solve the following biquadratic equation:<br><br> <img src="https://tex.z-dn.net/?f=%28%287%21%29%2Ax%5E4%29%2B%28%285%21%29%2Ax

jekas [21]

$\large\begin{array}{l} \textsf{Solve the equation}\\\\ \mathsf{7!\cdot x^4+5!\cdot x^2-3=0}\\\\ \mathsf{7!\cdot (x^2)^2+5!\cdot x^2-3=0}\\\\\\ \textsf{Substitute}\\\\ \mathsf{x^2=t\quad(t\ge 0)}\\\\\\ \textsf{so the equation becomes}\\\\ \mathsf{7!\cdot t^2+5!\cdot t-3=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=7!}\\\mathsf{b=5!}\\\mathsf{c=-3} \end{array} \right. \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(5!)^2-4\cdot 7!\cdot (-3)}\\\\ \mathsf{\Delta=(5!)^2+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7\cdot 6\cdot 5!}\\\\ \mathsf{\Delta=5!\cdot (5!+12\cdot 7\cdot 6)}\\\\ \mathsf{\Delta=5!\cdot (120+504)}\\\\ \mathsf{\Delta=5!\cdot 624} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=120\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3\cdot 5)\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^{3+4}\cdot 3\cdot 5 \cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^7\cdot 3^2\cdot 5 \cdot 13} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=2^{6+1}\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{6}\cdot 2\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{3\,\cdot\,2}\cdot 3^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 130} \end{array}$

$\large\begin{array}{l} \mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-(5!)\pm \sqrt{(2^3\cdot 3)^2\cdot 130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-120\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-2^3\cdot 3\cdot 5\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 4\cdot 7\cdot (2\cdot 3)\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 2^2\cdot 3\cdot 7\cdot 2\cdot 5\cdot 3!}} \end{array}$

$\large\begin{array}{l} \mathsf{t=\dfrac{(2^3\cdot 3)\cdot \big(\!\!-5\pm \sqrt{130}\big)}{(2^3\cdot 3)\cdot 7\cdot 2\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{7\cdot 2\cdot 5\cdot 3\cdot 2\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 7\cdot 5\cdot 3\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 105}} \end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}&~\textsf{ or }~& \end{array} \mathsf{t=\dfrac{-5-\sqrt{130}}{2^2\cdot 105}}\quad\textsf{(useless, since }\mathsf{t\ge 0}\textsf{)}\\\\ \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}} \end{array}$

$\large\begin{array}{l} \textsf{Substituite back for }\mathsf{x^2=t:}\\\\ \mathsf{x^2=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}\\\\ \mathsf{x=\pm \sqrt{\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}}\\\\ \mathsf{x=\pm \dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}\\\\ \boxed{\begin{array}{rcl} \mathsf{x=-\,\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}&~\textsf{ or }~&\mathsf{x=\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}} \end{array}} \end{array}$

$\large\begin{array}{l} \textsf{Solution: }\mathsf{S=\left\{-\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}},\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}}\right\}.} \end{array}$

If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2094277

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}$

Tags: <em>solve biquadratic equation factorial Bhaskara solution</em>

4 0

3 years ago