To justify that triangle CZB = triangle <span>CXA</span><span>, we have to use the AAS</span> congruence theorem.
If two angles and a non-included side of one triangle are congruent to the corresponding angles and side of another triangle, then the triangles are congruent.
Answer:
0.440559
Step-by-step explanation:
To find the probability of having a soldier from Iraq that had lung problems, we multiply the 2 probabilities, so 0.36*0.14 = 0.0504
The probability of a soldier serving not in Iraq and having lung problems is 0.64 (serving in Iraq is 0.36, so not serving in Iraq is 1 - 0.36 = 0.64) times 0.1 = 0.064.
If we sum up these two probabilities, we have the probability of a soldier having lung problems (0.0504+0.064 = 0.1144).
Now, to know the probability of a soldier with a lung problem having served on Iraq, we divide the probability of serving in Iraq and having lung problems (0.0504) by the probability of having lung problem (0.1144):
0.0504/0.1144 = 0.440559
Answer:
Theorem of calculus
Step-by-step explanation:
Take the integral

Integrate the sum term by term and factor out constants

Apply the fundamental theorem of calculus.
<em>The antiderivative of x^2 is x^3/3, while for a constat is x</em>

Evaluate the limits


Answer:
D. 7.5 inches
Step-by-step explanation:
So, we can find the ratio of the big triangle to the small one by dividing the values of the hypotenuse.
So, 18/6 = 3
Big:small = 3:1
So, using this we can make an equation to find AD.
Let's say ED=x.
AD is basically ED (x) + AE (5).
Since ED is part of the small triangle, and AD is part of the big one, we can use the 3:1 ratio from above.
So,
=3.
Multiply both sides by x, getting x+5 = 3x.
Subtract the x: -x -x
And you get 5=2x, giving you 2.5 = x.
However, we're not done yet, as 2.5 is the value of ED, but we need AD.
Remember, AD is = ED + AE.
So, AD = 2.5 + 5.
Finally, AD = 7.5
Answer:
A, B, C
Step-by-step explanation:
The top right corner is the positive side of the graph. If it goes below the X axis or to left of the Y axis, it is negative. (thus, the negative numbers)