Radius is 5, diameter is 10
because 4/3 x 5^3 pi = 500/3 pi cm^3
Math requires a lot of thing and a procedure of steps for each topic. Yes I agree
Answer: (7, 20)
Concept:
There are three general ways to solve systems of equations:
- Elimination
- Substitution
- Graphing
Since the question has specific requirements, we are going to use <u>substitution </u>to solve the equations.
Solve:
<u>Given equations</u>
y = 3x - 1
2x + 6 = y
<u>Substitute the y value since both equations has isolated [y]</u>
2x + 6 = 3x - 1
<u>Add 1 on both sides</u>
2x + 6 + 1 = 3x - 1 + 1
2x + 7 = 3x
<u>Subtract 2x on both sides</u>
2x + 7 - 2x = 3x - 2x
<u>Find the value of y</u>
y = 3x - 1
y = 3(7) - 1
y = 21 - 1
Hope this helps!! :)
Please let me know if you have any questions
Answer:
nElements = 0;
for (int i = 0; i < a2d.length; i++)
nElements += a2d[i].length;
isRectangular = true;
for (int i = 1; i < a2d.length && isRectangular; i++)
if (a2d[i].length != a2d[0].length)
isRectangular = false;
Step-by-step explanation:
2 dimensional array is created and initialize element with 0. and the taking loop for 2 dimensional array .such as
n Elements = 0;
for (int i = 0; i < a2d.length; i++)
n Elements += a2d[i].length;
A boolean variable isRectangular and 2 dimensional array has been created if rectangle is true the take a2d.length and isRectangular less than i and intiliaze i with 1 then increment the i .It will give true value.
If rectangular is false then apply if statement and see if a2d[i] is not equal to a2d[0]
if (a2d[i].length != a2d[0].length)
isRectangular = false;
