Please consider the attached diagram.
We have been given that angle DAB and angle DCB are right angles. We are also told that measure of angle DBC is 42 degrees. We are asked to find the measure of arc CAB.
We can see that angle DBC is inscribed angle of arc DC, so measure of arc DC is two times the measure of angle DBC.
We can also see that segment DB is diameter of given circle as inscribed angle of diameter is a right angle.
Since segment DB is diameter of given circle, so measure of arc DAB would be equal to half the measure of 360 degrees.
Therefore, the measure of arc CAB is 264 degrees.
A: 14/16 that is the answer
the number inside the parenthesis includes the interest rate ( annual percentage
looking at the possible functions, if you raise the interest rate to the power of 12, you would need to divide the number of years (t) by 12 as well
so the function could be rewritten as A=1000(1.023^12)^t/12
since the annual percentage rate is 2.3% ( 0.023)
divide 0.023/12*100 = 0.19% per month interest
In both cases there are more than one possible function sutisfying given data.
1. If
- x‑intercepts are (–5, 0), (2, 0), and (6, 0);
- the domain is –5 ≤ x ≤ 7;
- the range is –4 ≤ y ≤ 10,
then (see attached diagram for details) you can build infinetely many functions. From the diagram you can see two graphs: first - blue graph, second - red graph. Translating their maximum and minimum left and right you can obtain another function that satisfies the conditions above.
2. If
- x‑intercepts are (–4, 0) and (2, 0);
- the domain is all real numbers;
- the range is y ≥ –8,
then you can also build infinetely many functions. From the diagram you can see two graphs: first - blue graph, second - red graph. Translating their minimum left and right you can obtain another function that satisfies the conditions above.
Note, that these examples are not unique, you can draw a lot of different graphs of the functions.
Answer: yes, there are more than one possible function
Answer:
Step-by-step explanation:
Let 
and 
be the production level of milk and white chocolate-covered strawberries respectively. According to the given data, we know the total profit will be
The restrictions can be written as
All the restrictions can be plotted in the same graph to find the feasible region where all of them are met. The graph is shown in the image below
The optimal solution will be the level of production such that
* All restrictions are met
* The total profit is maximum
The optimal level of production can be found in (at least) one of the vertices of the feasible region. We'll try each one as follows
P(0,0)=0
P(400,200)=$2.25 (400)+$2.50 (200) = $1400
P(600,200)=$2.25 (600)+$2.50 (200) = $1850
P(800,0)=$2.25 (800)+$2.50 (0) = $1800
We must produce 600 milk chocolate-covered strawberries and 200 white chocolate-covered strawberries to have a maximum profit of $1850/month
