Answer:
![PS=2\sqrt{6}](https://tex.z-dn.net/?f=PS%3D2%5Csqrt%7B6%7D)
Step-by-step explanation:
we are given triangle as
we know that
triangle PRS and triangle PQS are similar
so, the ratios of their sides must be equal
we get
![\frac{RS}{PS}=\frac{PS}{QS}](https://tex.z-dn.net/?f=%5Cfrac%7BRS%7D%7BPS%7D%3D%5Cfrac%7BPS%7D%7BQS%7D)
now, we are given
RS=4
QS=6
we can plug it
![\frac{4}{PS}=\frac{PS}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7BPS%7D%3D%5Cfrac%7BPS%7D%7B6%7D)
![PS^2=6\times 4](https://tex.z-dn.net/?f=PS%5E2%3D6%5Ctimes%204)
![PS^2=24](https://tex.z-dn.net/?f=PS%5E2%3D24)
now, we can solve for PS
![PS=2\sqrt{6}](https://tex.z-dn.net/?f=PS%3D2%5Csqrt%7B6%7D)
Answer:
SA = 748π in²
General Formulas and Concepts:
<u>Symbols</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Geometry</u>
Surface Area of a Cylinder Formula: SA = 2πrh + 2πr²
- <em>r</em> is radius
- <em>h</em> is height
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
<em>r</em> = 11 in
<em>h</em> = 23 in
<u>Step 2: Find Surface Area</u>
- Substitute in variables [Surface Area of a Cylinder Formula]: SA = 2π(11 in)(23 in) + 2π(11 in)²
- Evaluate exponents: SA = 2π(11 in)(23 in) + 2π(121 in²)
- Multiply: SA = 506π in² + 242π in²
- Add: SA = 748π in²
<u>Q</u><u>uest</u><u>ion</u><u>:</u>
To Simplify:
118 {121÷(11×11)-(-4)-(3-7)}
<u>Solu</u><u>tion</u>:
↠118 {121÷121+4-(-4)}
↠118 {1+4+4}
↠118 {5+4}
↠118{9}
↠118×9
↠1062
▬▬▬▬▬▬▬▬▬▬▬▬
The upper Solution is done by applying BODMAS
<u>Abou</u><u>t</u><u> </u><u>BODMAS</u><u>:</u>
B→ Bracket
O→ Of
D→ Division
M→ Multiplication
A→ Addition
S→ Subtraction
Answer:
48
Step-by-step explanation:
2 x 2 x 2 x 2 = 16
16 + ?
4 x 4 x 4 = 64
0.5 x 64 = 32
32 + 16 = 48