Answer:
The correct option is;
∠AQS ≅ ∠BQS when AS = BS
Step-by-step explanation:
Given that AQ is equal to BQ. When AS is drawn congruent to BS, we have;
QS is congruent to SQ by reflective property
Therefore;
The three sides of triangle QAS are congruent to the three sides of triangle QBS, from which we have;
∠AQS and ∠BQS are corresponding angles, therefore;
∠AQS ≅∠BQS because corresponding angles of congruent triangles are also congruent.
Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
100
Step-by-step explanation:
Answer:
The solutions of the equation are 0 and 0.75.
Step-by-step explanation:
Given : Equation 
To find : All solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically ?
Solution :
Equation 

Either
or 
When
When 
Solve by quadratic formula, 





The solutions of the equation are 0 and 0.75.
For verification,
In the graph where the curve cut x-axis is the solution of the equation.
Refer the attached figure below.
Answer:
Size 50
Step-by-step explanation:
You should choose the larger sample size. This is because if you take the mean, it will help to cancel out outliers. For instance, most people exercise half an hour. With sample size of 10, there might be one person who doesn't exercise, thus dragging down the mean. A larger sample size would have a greater ratio of "average" people.