Given that we assume that all the bases of the triangles are parallel.
We can use AAA or Angle-Angle-Angle to prove that these triangles are similar.
Each parallel line creates the same angle when intersecting with the same side.
For example:
The bases of each triangle cross the left side of all the triangle.
Each angle made by the intersecting of the the parallel base and the side are the same.
Thus, each corresponding angle of all the triangles are congruent.
If these angles are congruent, then we have similar triangles.
Answer:
3 in²
Step-by-step explanation:
The fraction 9/16 is equal to the equation 9 ÷ 16.
9 ÷ 16 = 0.5652
Lets find the square root of 0.5652
So the length of one side is 0.75 in²
Now we'll multiply this by 4, for the length of the 4 equal sides.
0.75 x 4 = 3.
This means the perimeter is 3 in².
If the roots to such a polynomial are 2 and
, then we can write it as
courtesy of the fundamental theorem of algebra. Now expanding yields
which would be the correct answer, but clearly this option is not listed. Which is silly, because none of the offered solutions are *the* polynomial of lowest degree and leading coefficient 1.
So this makes me think you're expected to increase the multiplicity of one of the given roots, or you're expected to pull another root out of thin air. Judging by the choices, I think it's the latter, and that you're somehow supposed to know to use
as a root. In this case, that would make our polynomial
so that the answer is (probably) the third choice.
Whoever originally wrote this question should reevaluate their word choice...
Assuming that the two investments are X & Y
X + Y = 6300
X = 6300 - Y (1)
9/100X + 4/100 Y = 372 (2)
replacing X from (1) into (2)
9/100(6300-Y) + 4/100 Y = 372
567 - 9/100Y +4/100Y = 372
(-9+4)/100Y = 372 - 567
5/100Y = -195
Y = 100*195/5 = 3900
From (1) we can get X
X = 6300 - 3900 = 2400
I hope this is helpful