It would be 42,523 because 50x10=500 and 5 in 42,523 is in the hundreds spot
Answer:
2299 in^2
Step-by-step explanation:
The formula for the volume of a square pyramid of base A and height h is
V = (A)(h).
Here, the perimeter of the base is 44 inches, so that the length of one side must be (44 in) / 4, or 11 in. The area of a square of side s is A = s^2, so we have:
Here, V = (11 in)^2 * (19 in) = 2299 in^2
Answer:
work is shown and pictured
The required coordinates of the unit circle for s = -11π/6 is (√3/2, 1/2). Using trigonometric functions, the required answer is calculated.
<h3>What are the trigonometric functions?</h3>
There are six trigonometric functions.
They are sin θ, cos θ, tan θ, sec θ, cosec θ, and cot θ.
Where θ - 0 to 360° or 0 to 2π
<h3>Calculation:</h3>
For a unit circle, the coordinates are given in the form of trigonometric functions as below:
x = cos θ and y = sin θ
Here it is given that s = θ = -11π/6
So, on substituting,
x = cos θ = cos (-11π/6)
⇒ x = √3/2
y = sin θ = sin (-11π/6)
⇒ y = 1/2
Thus, the required coordinate pair is (√3/2, 1/2).
Learn more about trigonometric functions here:
brainly.com/question/2291993
#SPJ1
Answer:
A two-digit number can be written as:
a*10 + b*1
Where a and b are single-digit numbers, and a ≠ 0.
We know that:
"The sum of a two-digit number and the number obtained by interchanging the digits is 132."
then:
a*10 + b*1 + (b*10 + a*1) = 132
And we also know that the digits differ by 2.
then:
a = b + 2
or
a = b - 2
So let's solve this:
We start with the equation:
a*10 + b*1 + (b*10 + a*1) = 132
(a*10 + a) + (b*10 + b) = 132
a*11 + b*11 = 132
(a + b)*11 = 132
(a + b) = 132/11 = 12
Then:
a + b = 12
And remember that:
a = b + 2
or
a = b - 2
Then if we select the first one, we get:
a + b = 12
(b + 2) + b = 12
2*b + 2 = 12
2*b = 12 -2 = 10
b = 10/2 = 5
b = 5
then a = b + 2= 5 + 2 = 7
The number is 75.
And if we selected:
a = b - 2, we would get the number 57.
Both are valid solutions because we are changing the order of the digits, so is the same:
75 + 57
than
57 + 75.
