- Given - <u>an </u><u>equation</u><u> </u><u>in </u><u>a </u><u>standard</u><u> </u><u>form</u>
- To do - <u>simplify</u><u> </u><u>the </u><u>equation</u><u> </u><u>so </u><u>as </u><u>to </u><u>obtain </u><u>an </u><u>easier </u><u>one</u>
<u>Since </u><u>the </u><u>equation</u><u> </u><u>provided </u><u>isn't</u><u> </u><u>i</u><u>n</u><u> </u><u>it's</u><u> </u><u>general</u><u> </u><u>form </u><u>,</u><u> </u><u>let's</u><u> </u><u>first </u><u>convert </u><u>it </u><u>~</u>
<u>General</u><u> </u><u>form </u><u>of </u><u>a </u><u>Linear</u><u> equation</u><u> </u><u>-</u>
<u>T</u><u>he </u><u>equation</u><u> </u><u>after </u><u>getting</u><u> </u><u>converted</u><u> </u><u>will </u><u>be </u><u>as </u><u>follows</u><u> </u><u>~</u>
hope helpful ~
Answer:
Cos A = (√51)/ 10
Step-by-step explanation:
sin is opposite/hypothenuse
cos is adjacent/hypothenuse
Pythagorus theorem says that c² = a² + b²
10² = 7² + b²
b² = 100 - 49
b = ±√51
It doesn't make sense for a length to be a minus number therefore, we will use +√51.
Cos A = (√51)/ 10
Tell me if I am wrong.
Can I get brainliest
The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.
Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have
The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have
The second sum is simply the sum of N ones:
So, the final result is
which ends the proof.
Answer:
TWO BUCKS?? ballin on a budget
anyways, if there are 16 guests, the party would cost $50. sounds a lot more correct lol