The area of the shaded triangle formed as the result of the overlap is = 62.35 inches ²
<h3>Calculation of the equilateral triangle</h3>
After folding the rectangle with length of 12 inches and width of 18 inches, an equilateral triangle was formed.
An equilateral triangle is a type of triangle where by all the three sides are equal.
To determine the value of one of the sides, CB or CD is used because the folding didn't affect these sides.
Using the formula for the area of an equilateral triangle,
A = √¾ a²
a= 12 inches
A = √¾ ×12²
A = 62.35 inches ²
Learn more about triangle here:
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<u>Given</u>
AC = BC
<u>Substitution</u>
∠1 = ∠3
<u>Base ∠'s of isosceles triangle =</u>
∠1 = ∠3
<u>Vertical angles =</u>
∠2 = ∠3
<h3>Answers are:
sine, tangent, cosecant, cotangent</h3>
Explanation:
On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.
Since sine is negative, so is cosecant as this is the reciprocal of sine
csc = 1/sin
In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.
Because tangent is negative, so is cotangent.
The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.
x is the temperature and y is the number of swimmers. If there are 80 swimmers, then y = 80 instead of x = 80
This is how Corey should have solved
y = 1.505*x - 88.21
80 = 1.505*x - 88.21 <--- replace y with 80
1.505*x - 88.21 = 80
1.505*x = 80 + 88.21 <--- add 88.21 to both sides
1.505*x = 168.21
x = 168.21/1.505 <--- divide both sides by 1.505
x = 111.767
x = 111.8
If there are 80 swimmers at the pool, then the possible outside temperature is roughly 111.8 degrees F. This temperature seems awfully large, so it's possible that Corey's regression equation has limited scope; in other words, it isn't too useful if x is outside a given interval
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