I believe 100mg. I'm not so sure though but, you can always check google you know..
Answer:
40.56
Step-by-step explanation:
you need the circumference of the cirle aka perimeter of circle so 8(pi) and then because its only half, divided by 2... 12.56
now you hgave circumference so add 12.56 to 28 and there you go. you can round how ever necessary
Answer:
16
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Trigonometry</u>
[Right Triangles Only] Pythagorean Theorem: a² + b² = c²
- a is a leg
- b is another leg
- c is the hypotenuse<u>
</u>
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
Leg <em>a</em> = <em>a</em>
Leg <em>b</em> = 12
Hypotenuse <em>c</em> = 20
<u>Step 2: Solve for </u><em><u>a</u></em>
- Substitute in variables [Pythagorean Theorem]: a² + 12² = 20²
- Evaluate exponents: a² + 144 = 400
- [Subtraction Property of Equality] Isolate <em>a</em> term: a² = 256
- [Equality Property] Square root both sides: a = 16
It’s a ! you add both of the numbers to get your missing perimeter.
I’m very sorry if it’s wrong but this is what I believe! stay safe
Answer:
Cos A=5/13
we have
Cos² A=
25/169=1-Sin²A
sin²A=1-25/169
sin²A=144/169
Sin A=
again
Tan B=4/3
P/b=4/3
p=4
b=3
h=
Now
Sin B=p/h=4/5
in IV quadrant sin angle is negative so
Sin B=-4/5
CosB=b/h=3/5
Now
<u>S</u><u>i</u><u>n</u><u>(</u><u>A</u><u>+</u><u>B</u><u>)</u><u>:</u><u>s</u><u>i</u><u>n</u><u>A</u><u>c</u><u>o</u><u>s</u><u>B</u><u>+</u><u>C</u><u>o</u><u>s</u><u>A</u><u>s</u><u>i</u><u>n</u><u>B</u>
<u>n</u><u>o</u><u>w</u><u> </u>
<u>substitute</u><u> </u><u>value</u>
<u>Sin(A+B):</u>12/13*3/5+5/13*(-4/5)=36/65-4/13
<u>=</u><u>1</u><u>6</u><u>/</u><u>6</u><u>5</u><u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u>