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Alex777 [14]
3 years ago
11

???!! Someone pleaseee help me

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
8 0
If AB + BC = AC, and A, B and C are collinear, then
B is between A and C
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Given the following kite find:
arsen [322]

Answer:

50 degrees

Step-by-step explanation:

in a kite the diagonals are perpendicular and the smallest one divide it in two isosceles triangles. This mean that if the see at the top one we have that the two angles of its base are both 40 degree

a triangle has the sum of interior angles that is 180 degrees

180 - 80 = 100 degrees

the height of a isosceles triangle is also its bisector

a bisector divides an angle in two congruent angles, so we have

1: 100/2 = 50 degrees

5 0
3 years ago
Read 2 more answers
Someone please help Austin writes down a random number. What is the probability that the number he wrote down is an odd number A
mylen [45]

Answer:

D

Step-by-step explanation:

every number is either even or odd so half of infinite possibilities.

6 0
3 years ago
The ravens won three more games than they lost. they played 96 games. how many games did they win
ololo11 [35]

divide 96 by 2  which gives you 48 and subtract 3 since they lost 3 which is 45. you the subtract 96 from 45 which gives you 51 which is the answer

6 0
4 years ago
Which ordered pair is a solution to the system of inequalities?
gladu [14]

Answer: The answer is C. Hope this helps :)

Step-by-step explanation:

3 0
3 years ago
Determine an expression for dy<br> dx given that x = sin3<br> (t)andy = cos3<br> (t)
shtirl [24]

Answer:

\frac{dy}{dx} =-\sqrt[3]{\frac{y}{x} }

Step-by-step explanation:

Recall that using the chain rule we can state:

\frac{dy}{dt} =\frac{dy}{dx}*\frac{dx}{dt}

and therefore solve for dy/dx as long as dx/dt is different from zero.

Then we find dy/dt  and dx/dt,

Given that

x=sin^3(t)\\dx/dt = 3 sin^2(t)* cos(t)

And similarly:

y=cos^3(t)\\dy/dt=-3\,cos^2(t)*sin(t)

Therefore, dy/dx can be determined by the quotient of the expressions we just found:

\frac{dy}{dx} =\frac{dy/dt}{dx/dt} =\frac{-3\,cos^2(t)*sin(t)}{3\,sin^2(t)*cos(t)} =-\frac{cos(t)}{sin(t)}

now notice that we can find   cos(t) = \sqrt[3]{y}  from the expression for y,

and   sin(t) = \sqrt[3]{x}  from its expression for x.

Therefore dy/dx can be written in terms of x and y as:

\frac{dy}{dx} =-\frac{cos(t)}{sin(t)}=-\sqrt[3]{\frac{y}{x} }

4 0
3 years ago
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