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3 years ago

How do you solve a SSA triangle?

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

5 0

Since sin(100)/5 is equivalent to sinA/9, then:
$sinA = \frac{9sin(100)}{5}$
$A = sin^{-1}(\frac{9sin(100)}{5}$

Type that into your calc to find A.

Repeat this process with C.

You might be interested in

Solve the absolute value equation |x+4|=|7-x|

Artemon [7]

Answer:

x+7=7-x or

=-(7-x)

x=3/2 is the only solution

8 0

3 years ago

What is the solution to negative -33+ M=10?

Novosadov [1.4K]

The solution is 43.

5 0

3 years ago

Read 2 more answers

)If 18th February, 2030 falls on Monday then what will be the day on 18th February, 2040

MariettaO [177]

The Saturday will be the day on 18th February 2040 if 18th February 2030 falls on Monday

<h3>What is an arithmetic operation?</h3>

It is defined as the operation in which we do the addition of numbers, subtraction, multiplication, and division. It has a basic four operators that is +, -, ×, and ÷.

We have:

18th February 2030 falls on Monday

To find what will be the day on 18th February 2040

Calculate the expression:

= Given date + Month code + (difference between years) + Numbe of leap year

Leap year = difference between years/4

= (2040-2030)/4

= 10/4 = 2.5 ≈2

= 18 + 3 + (2040-2030) + 2

= 33

Divide 33 by 7 the remainder will be 5

Day code = given day code + remainder

= 1 + 5

= 6 (saturday from the table attahced)

Thus, Saturday will be the day on 18th February 2040 if 18th February 2030 falls on Monday.

Learn more about the arithmetic operation here:

brainly.com/question/20595275

#SPJ1

6 0

2 years ago

0.83 compared 83/100

GarryVolchara [31]

" 0.83 " MEANS " 83/100 ". They are identical quantities, because they are the SAME quantity.

4 0

3 years ago

Pls solve this question

just olya [345]

Answer:

${( \sqrt{ {x}^{ - 3} }) }^{5} = ({( {x}^{ - 3}) }^{ \frac{1}{2} } ) ^{5} \\ \\ = {x}^{ - 3 \times \frac{1}{2} \times 5} = {x}^{ - \frac{15}{2} } = \frac{1}{ {x}^{ \frac{15}{2} } }$

<em><u>Or ;</u></em>

${( \sqrt{ {x}^{ - 3} } )}^{5} = {( \sqrt{ \frac{1}{ {x}^{3} }} })^{5} = ( { \frac{ \sqrt{1} }{ \sqrt{ {x}^{3} } } })^{5} \\ \\ = ( { \frac{1}{ \sqrt{x} \sqrt{ {x}^{2} } } })^{5} = ({ \frac{1}{ x\sqrt{x} }})^{5} \\ \\ = ( \frac{ {(1)}^{5} }{ {x}^{5} ( \sqrt{x} )^{5} } ) = \frac{1}{ {x}^{5} \times {x}^{ \frac{1}{2} \times 5 } } \\ \\ = \frac{1}{ {x}^{5} \times {x}^{ \frac{5}{2} } } = \frac{1}{ {x}^{5} \sqrt{ {x}^{5} } } \\ \\ = \frac{1}{ {x}^{5} \sqrt{ {x}^{4} {x}^{1} } } = \frac{1}{ {x}^{5} \sqrt{x} \sqrt{( { {x}^{2} })^{2} } } \\ \\ = \frac{1}{ {x}^{5} {x}^{2} \sqrt{x} } = \frac{1}{ {x}^{7} \sqrt{x} } = \frac{ \sqrt{x} }{ {x}^{8} }$

3 0

2 years ago