Answer:
Step-by-step explanation:
product of slopes of perpendicular lines=-1
(t-5)/(3+4)×(2-3)/(-4-1)=-1
(t-5)/7×(-1/-5)=-1
(t-5)/35=-1
t-5=-1×35=-35
t=-35+5
t=-30
2.
slopes of parallel lines are equal.
(-2+3)/(t-4)=(-1-4)/(4+2)
1/(t-4)=-5/6
t-4=-6/5
t=4-6/5=(20-6)/5=14/5
3.
x>0,y<0
so P lies in4th quadrant.
except cos and sec all are negative.
so only cos and sec are positive.
Look at the image below where I labeled the sides
To solve this you must use Pythagorean theorem:
a and b are the legs (the sides that form a perpendicular/right angle)
c is the hypotenuse (the side opposite the right angle)
In this case...
a = 4
b = 6
c = unknown
^^^Plug these numbers into the theorem
simplify
16 + 36 = 
52 =
To remove the square from x take the square root of both sides to get you...
√52 = x
^^^Unsimplified radical
2√13
^^^Simplified radical
7.21
^^^Rounded to hundedths
Hope this helped!
~Just a girl in love with Shawn Mendes
Answer:
∠1 is 33°
∠2 is 57°
∠3 is 57°
∠4 is 33°
Step-by-step explanation:
First off, we already know that ∠2 is 57° because of alternate interior angles.
Second, it's important to know that rhombus' diagonals bisect each other; meaning they form 90° angles in the intersection. Another cool thing is that the diagonals bisect the existing angles in the rhombus. Therefore, 57° is just half of something.
Then, you basically just do some other pain-in-the-butt things after.
Since that ∠2 is just the bisected half from one existing angle, that means that ∠3 is just the other half; meaning that ∠3 is 57°, as well.
Next is to just find the missing angle ∠1. Since we already know ∠3 is 57°, we can just add that to the 90° that the diagonals formed at the intersection.
57° + 90° = 147°
180° - 147° = 33°
∠1 is 33°
Finally, since that ∠4 is just an alternate interior angle of ∠1, ∠4 is 33°, too.
Answer:
5
Step-by-step explanation:
The side with 4 cubes has a length of 2, the side with 2 cubes has a length of 1, and the side with 5 cubes has a length of 2.5.
2 x 1 x 2.5 = 5
