Answer:
The answer is "Option a"
Explanation:
Range-based for loop performs a sequence for a loop. It's more accessible as the conventional loop, for example, all components in the array, running more than a range of possibilities. In the given question "option a" is correct because it follows the correct syntax and other choices were wrong, which can be described as follows:
- In option b, It's not correct, because in this code the range declaration is wrong.
- In option c, It is wrong, because in this code the datatype is missing.
- In option d, It is illegal syntax, that's why it is wrong.
Answer:
1, 4, 7
Explanation:
The instruction in the question can be represented as:
for i in range(1,10,3):
print i
What the above code does is that:
It starts printing the value of i from 1
Increment by 3
Then stop printing at 9 (i.e.. 10 - 1)
So: The sequence is as follows
Print 1
Add 3, to give 4
Print 4
Add 3, to give 7
Print 7
Add 3, to give 10 (10 > 10 - 1).
So, it stops execution.
Answer:An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.
The video above uses the example
{
d
y
d
x
=
cos
(
x
)
y
(
0
)
=
−
1
to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you
y
=
sin
(
x
)
+
C
.
Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in
0
for
x
and
−
1
for
y
gives us
−
1
=
C
, meaning that the particular solution must be
y
=
sin
(
x
)
−
1
.
So the general way to solve initial value problems is: - First, find the general solution while ignoring the initial condition. - Then, use the initial condition to plug in values and find a particular solution.
Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be
y
-values. Higher-order equations might have an initial value for both
y
and
y
′
, for example.
Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.
Explanation:
Answer:
See attachment for flowchart
Explanation:
The flowchart is represented by the following algorithm:
1. Start
2. Input Mass
3. Input Volume
4 Density = Mass/Volume
5. Print Density
6. Stop
The flowchart is explained by the algorithm above.
It starts by accepting input for Mass
Then it accepts input for Volume
Step 4 of the flowchart/algorithm calculated the Density using the following formula: Density = Mass/Volume
Step 5 prints the calculated Density
The flowchart stops execution afterwards
Note that the flowchart assumes that the user input is of number type (integer, float, double, etc.)
