3 2/3 x # = 2/9
# = 2/9 \ 3 2/3
# = 2/33 or 0.067
Answer:
x = 70°
Step-by-step explanation:
The relevant relations are ...
- base angles of an isosceles triangle are congruent
- consecutive interior angles where parallel lines meet a transversal are supplementary
__
Triangle OBQ is isosceles, so angle OQB = 40°. Triangle OPQ is isosceles, so angle OQP is x. The sum of angles OQB and OQP is angle BQP, which is supplementary to angle OPQ. That is, ...
(40° +x) +x = 180°
2x = 140° . . . . . . subtract 40°
x = 70° . . . . . . divide by 2
_____
There are many ways to find x. The one shown here is just one of them. In general, right triangles, isosceles triangles, symmetry, inscribed angles can all be used to write relations involving the known angles and x.
The function involves the initial speed of the object, and the height
from which it was launched, and whether it was launched upward
or downward. As you can see, your post was corrupted and is
unreadable. But when the object hits the ground, its height is zero.
Set the function equal to zero, solve the quadratic equation that you
have, discard the negative solution, and the positive solution is your
answer.
Use this website called mathpapa.com
It’s really helpful and completely free
Hope this helped!
Answer:
My answer:
Step-by-step explanation:
- -14K for the first blank
- -14K for the second blank
- -56 for the third blank
- 4 for the fourth blank
I multiplied -7 by 2. Then I plugged in that number to the first two blanks. Then I subtracted 21 from -35. Then I divided that number by -14.
