we have the following system of equations

we know that
The intersection of both graphs is the solution of the system of equations
using a graph tool
see the attached figure
the solution is the point 
therefore
<u>the answer is</u>
the solution of the system of equations is the point 
Answer:
Step-by-step explanation:
we know



(a)![\left [ \left ( \hat{i}+\hat{j}\right )\times \hat{i}\right ]\times \hat{j}](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cleft%20%28%20%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5Cright%20%29%5Ctimes%20%5Chat%7Bi%7D%5Cright%20%5D%5Ctimes%20%5Chat%7Bj%7D)
![=\left [ \hat{i}\times \hat{i}+\hat{j}\times \hat{i}\right ]\times \hat{j}](https://tex.z-dn.net/?f=%3D%5Cleft%20%5B%20%5Chat%7Bi%7D%5Ctimes%20%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5Ctimes%20%5Chat%7Bi%7D%5Cright%20%5D%5Ctimes%20%5Chat%7Bj%7D)
![=\left [ 0-\hat{k}\right ]\times \hat{j}](https://tex.z-dn.net/?f=%3D%5Cleft%20%5B%200-%5Chat%7Bk%7D%5Cright%20%5D%5Ctimes%20%5Chat%7Bj%7D)

(b)


(c)


(d)



Answer:
26.11% of the test scores during the past year exceeded 83.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that
.
Approximately what percentatge of the test scores during the past year exceeded 83?
This is 1 subtracted by the pvalue of Z when
. So:



has a pvalue of 0.7389.
This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.
Answer:
Value of v that minimizes E is v = 3u/2
Step-by-step explanation:
We are given that;
E(v) = av³L/(v-u)
Now, using the quotient rule, we have;
dE/dv = [(v-u)•3av²L - av³L(1)]/(v - u)²
Expanding and equating to zero, we have;
[3av³L - 3av²uL - av³L]/(v - u)² = 0
This gives;
(2av³L - 3av²uL)/(v-u)² = 0
Multiply both sides by (v-u)² to give;
(2av³L - 3av²uL) = 0
Thus, 2av³L = 3av²uL
Like terms cancel to give;
2v = 3u
Thus, v = 3u/2
Tree...........................dock
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boat
use Pythagorean Theorem to satisfy...a^2 + b^2 = c^2
200^2 + b^2 = 300^2
b^2 = 300^2 - 200^2
b^2 = 50000 (get square root of each)
b = 223.606 (rounded to the nearest tenth.... = 223.6