Answer:
BD=46
Step-by-step explanation:
In the given question we have a <em>Triangle Δ BCD</em>, that is intersected by CA, and dividing Δ BCD in half. The line CA creates two new Right Triangles denoted as Δ BCA and Δ DCA.
Right Triangles are characterized by one orthogonal angle that is always 90° and two more angles (lets call them ∠x and ∠y) so that:
∠x + ∠y + ∠90° = 180°
In this case we can start analysing each Right triangle individually at first. As such, we shall use the Pythagorian Theorem which relates all three sides of a Right triangle as follow:
<u>In Δ BCA:</u>
Eqn(1).
<u>In Δ DCA:</u>
Eqn(2).
<em>which means that the Hypotenuse Length Squared (in this case BC and CD) are equal to the sum of the remaining two sides length's squared (in this case AB and AC for Δ BCA and similarly AD and AC for Δ DCA)</em>
In the question it is also given that BC = CD which implies that Eqn(1) = Eqn(2) as follow:
Gathering similar sides/terms
AC cancels out
Transfering AD on the right hand side
AB=AD
We also know that 
, hence from our result above we can say that
Eqn(3)
Finally since 
, as shown on the Schematic Attached and along with Eqn(3) we have
<em>There are other methods of proving such result using Pythagorian Theorem, Angles Relations and/or Triangle Sides/Comparison. </em>
2.5
2.33..repeating
4.5
3.33...repeating
the repeating means there is an infinite number of 3's after the decimal
Answer:
a: down b: (1.25, 0.125) c: x=1.25
Step-by-step explanation:
Use a graphing calculator to see the graph and determine answers.
Answer:
672
Step-by-step explanation:
we multiply 28 by 24, as in each row there are 24 chairs, and there are 28 rows total
28 x 24 = 672
there were 672 chairs in total for graduation
<u>part b:</u>
we see that there are 672 chairs total for the graduation as shown in part a. we are then told that these 672 chairs were loaded into 4 trucks. to solve, we divide 672 by 4 to see how many chairs were in each truck:
672 / 4 = 168 < use long division to solve 672/4
Hello! So the low and high whiskers consists of 1 and 10. A has 1 in the data, but not 10. A is eliminated. C has 10 in the data, but does not have one. That eliminates answer choice C. B has 1 and 10 as the data. Let's put the data of the numbers in order and see what happens.
1, 3, 4, 5, 5, 8, 9, 9, 9, 10
Let's cross out numbers from each end of the line to find our median. When we do this, we see that 5 and 8 are the middle numbers. In this case, we add the numbers and divide by 2. 5 + 8 is 13. 13/2 is 6.5 There. The median in the data is 6.5, and the box-and-whisker plot shows 6.5 as the median. There's a chance that ti could be it. But let's solve for the lower quartile. Draw a line between the 5 and 8. Start of by crossing out the 1 and 5 on opposite sides on the half. We see that the lower quartile is 4. Repeat the same step for the opposite half to find the third quartile, and the third quartile is 9. 4 and 9 are both shown as the quartiles in the box and whisker plot. B shows the correct data for the plot shown above. The answer is B.
