The quastion is asking us to determine the measure of angle M in a right triangle LMN. We know that angle N is a right angle and that LM = 76 ( Hypotenuse ) and MN = 40 ( Adjacent ). We will use trigonometry: cos M = Adjacent / Hypotenuse = 40 / 76 = 0.5263158; M = cos^(-1) 0.563158; M = 58.242° = 58° 15`. Answer: The measure of angle M is<span> 58°15`.</span>
Answer:
150
Step-by-step explanation:
First you have to find what number 80 is 2/5 of, to do this do the reciprocal of 2/5 which would be 5/2. Multiply 80 by 5/2 and you'll get 200, then find what 3/4 of 200. Multiply 200 by 3, then divide by 4 to get the answer of 150.
Since you don't provide the coordinates of the point W, I will help you in a general form anyway. In the Figure below is represented the segment that matches this problem. We have two endpoints U and V. So, by using the midpoint formula we may solve this problem:
Therefore:
So we know 
but we also must know 
Finally, knowing the points U and W we can find the endpoint V.
Answer:
The factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....
Step-by-step explanation:
The given expression is:
2q²-5pq-2q+5p
Make a pair of first two terms and last two terms:
(2q²-5pq) - (2q-5p)
Now factor out the common factor from each group.
Note that there is no common factor in second group. So we will take 1 as a common factor.
q(2q-5p) -1(2q-5p)
Now factor the polynomial by factoring out the G.C.F, 2q-5p
(2q-5p) (q-1)
Thus the factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....
Answer:
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.
<u><em>Hope this helps!</em></u>
