Answer:
AB = 75
BC = 60
AC = 45
m∠A = 53°
m∠B = 37°
m∠C = 90°
Step-by-step explanation:
<u>Trigonometric ratios</u>
where:
is the angle- O is the side opposite the angle
- A is the side adjacent the angle
- H is the hypotenuse (the side opposite the right angle)
Given:
Therefore:
- side opposite angle A = BC = 60
- side adjacent angle A = AC = 45
To find the length of AB (the hypotenuse), use Pythagoras’ Theorem:
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
⇒ AC² + BC² = AB²
⇒ 45² + 60² = AB²
⇒ AB² = 5625
⇒ AB = √5625
⇒ AB = 75
To find m∠A:
m∠C = 90° (as it is a right angle)
The interior angles of a triangle sum to 180°
⇒ m∠A + m∠B + m∠C = 180°
⇒ 53° + m∠B + 90° = 180°
⇒ m∠B = 180° - 53° - 90°
⇒ m∠B = 37°
h has infinitely many solutions.
The <u>correct answer</u> is:
A) The medians are both between 10 and 14 emails.
Explanation:
The <u>mode </u>is the easiest measure to find of a data set.
The <u>mode </u>of a data set is the data value that appears the most often. In plot A, there are 3 dots at 10 and 3 dots at 15; this means the modes are 10 and 15.
In plot B, there are 3 dots at 5 and 3 dots at 15; this means the modes are 5 and 15.
They <u>do not have the same modes</u>.
The <u>median </u>of a data set is the middle value. There are 10 dots in each dot plot; this means the medians will each be between two data points.
For plot A, we can see that the middle value is between 10 and 15.
For plot B, we can see that the middle value is between 10 and 15.
This means that choice A is correct, the medians of both are between 10 and 14.
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
Omg my name is Samantha sorry i can’t help you. I hope someone can help you!!!!
