Answer:
500 bricks.
Step-by-step explanation:
There are 250 bricks used to build a wall that is 20 feet high how many bricks will be used to build a wall is 40 feet high?
To solve e the above question, we have:
20 feet high = 250 bricks
40 feet high = x bricks
Cross Multiply
20 feet × x bricks = 40 feet × 250 bricks
x bricks = 40 feet × 250 bricks/20 feet
x bricks = 500 bricks
Therefore, 500 bricks will be used to build a wall is 40 feet high.
Answer:
This statement is false. Exterior angles are the angles created when the sides of the triangle are extended.
Step-by-step explanation:
Exterior angles are the angle between the any reference side and line extended from the adjacent side.
the sum of exterior and interior angle is 180 as it lies on a straight line.
_______________________________________________
based on above definition the given statement becomes false.
Hence, correct choice is option third
This statement is false. Exterior angles are the angles created when the sides of the triangle are extended.
Five squared with the little two and 2 because 5 Times 10 is 50 and five is prime so then from 10 and you get five and two and both five and two are prime
Answer:
5/8
Step-by-step explanation:
1 inches in fraction form would be 8/8 inches, and the button is 3/8 inches, so a 1 inch button is 5/8 inches longer than a 3/8 inch button
Answer:

Step-by-step explanation:
A second order linear , homogeneous ordinary differential equation has form
.
Given: 
Let
be it's solution.
We get,

Since
, 
{ we know that for equation
, roots are of form
}
We get,

For two complex roots
, the general solution is of form 
i.e 
Applying conditions y(0)=1 on
, 
So, equation becomes 
On differentiating with respect to t, we get

Applying condition: y'(0)=0, we get 
Therefore,
