Answer:
159
Step-by-step explanation:
Based on process of elimination, it cant be 51 nor 55 because that angle is 100% bigger than 90 degrees. To find it correctly, all you would do is add them. 105 + 54= 159
Answer:
0.624999999(repeating) so i would round it to 0.625 or 0.63 (both work and are rounded, its really up to wherever your teacher wants you to round it to if they want you to round it)
Step-by-step explanation:
2/3 x 1/2 (3/4 ÷ 2/5)=
simplify the parentheses (3/4 ÷ 2/5) to 1.875
2/3 x 1/2 x 1.875=
simplify 2/3 x 1/2 to 0.333333(repeating)
0.3333333333 x 1.875 = 0.6249999999(repeating)
round it to 0.625 or 0.63 (which ever one your teacher wants you to round it to)
Answer:
x = 45
y = 45
z = 11.1
Step-by-step explanation:
Remark
The triangle is a right triangle. That means that it has 1 right angle.
It is also isosceles which means that the two legs that are not the hypotenuse are equal.
Result: z and 11.1 are equal z = 11.1
More remark
If the triangle is isosceles, that means that the angles that are not right angles must also be equal.
90 + x + y = 180
but x = y
90 + 2x = 180 Subtract 90
2x = 180 - 90 Divide by 2
2x/2 = 90/2
x = 45
Step-by-step explanation:
Let ABC be an isosceles triangle with sides AC and BC of equal length.
We need to prove that the medians AD and BE are of equal length.
Consider the triangles ADC and BEC.
They have two congruent sides that include congruent angles.
Indeed, AC = BC by the condition, because the triangle ABC is isosceles.
Since the lateral sides AC and BC are of equal length, their halves EC
and DC are of equal length too: EC = DC.
Finally, the angle ECD is the common angle.
Thus, the triangles ADC and BEC are congruent, in accordance to the
postulate P1 (SAS) (see the lesson Congruence tests for triangles of the
topic Triangles in the section Geometry in this site).
Hence, the medians AD and BE are of equal length as the corresponding sides
of these triangles.
The proof is completed.
Answer: 15120
Step-by-step explanation:
Given: The number of players at point guard = 10
The number of players at shooting guard = 12
The number of players at small forward = 7
The number of players at power forward = 2
The number of players at center = 9
Then, the number of different teams William could draft is given by :-
10∗12∗7∗2∗9=15120
Hence, William could draft 15120 different teams.
