Weird way to write it but alright! (Sideways)
19pq^-2 x 5pq^6 = ?
These problems are pretty much single operations between each of the variables / constants.
So it's like this:
(19*5)(p*p)(q^-2*q^6) = ?
19*5 is 95.
For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.
p*p is p^2
For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:
-2 + 6 = 4
(The two is negative, because the power is negative 2)
So, q^4.
Our final answer is all of the combined.... like a so:
95p^2q^4
That is technically impossible but I can put this into a mixed fraction which is -5 1/4
When we are given 3 sides, we try to solve the angles first by using the
law of cosines
cos (A) = [b^2 + c^2 - a^2] / (2 * b * c)
cos (A) = [43^2 + 17^2 -27^2] / (2 * 43 * 17)
cos (A) = [1,849 + 289 -729] /
<span>
<span>
<span>
1,462
</span></span></span>cos (A) = 1,409 / 1,462
cos (A) =
<span>
<span>
<span>
0.96374829001368
Angle A = 15.475
Now that we have one angle, we next can use the
Law of Sines
sin(B) / side b = sin(A) / side a
sin(B) = sin(A) * sideb / sidea
</span></span></span><span>sin(B) = sin(15.475) * 43 / 27
</span><span>sin(B) = 0.26682 * 43 / 27
sin (B) = </span><span>0.424935555555</span>
Angle B = 25.147 Degrees
Remember the arc sine (<span>0.424935555555) also equals </span>
<span>
<span>
<span>
154.85
</span></span></span>Finally, calculating the third angle is quite easy
Angle C = 180 - Angle (A) - Angle(B)
Angle C = 180 - 15.475 - 154.85
Angle C = 9.675
Source:
http://www.1728.org/trigtut2.htm
What grade is this? I can get the answer of it but I forgot how
Answer:
Answer and explanation with steps are in the following attachments
Step-by-step explanation: