Answer:
The answer to your question is Area = 39.15 in²
Step-by-step explanation:
Data
See the picture below
Process
1.- Calculate the area of rectangle 1
Area = base x height
Area = (8.8)(6.55 - 4.3)
Area = (8.8)(2.25)
Area = 19.8 in²
2.- Calculate the area of rectangle 2
Area = (4.3)(2.25)
Area = 9.675 in²
3.- Calculate the area of rectangle 3
Area = (4.3)(2.25)
Area = 9.675 in²
4.- Calculate the total area
Area = 19.8 + 9.675 + 9.675
Area = 39.15 in²
Second way
Area 1 = 2[6.55 x 2.25]
Area 1 = 29.475 in²
Area 2 = (8.8 - 2.25 - 2.25)(6.55 - 4.3)
Area 2 = 4.3 x 2.25
Area 2 = 9.675 in²
Total area = 29.475 + 9.675
= 39.15 in²
Third way
Area 1 = (8.8 - 2.25)(6.55 - 4.3)
= (6.55)(2.25)
= 14.7375 in²
Area 2 = (4.3)(2.25)
= 9.675 in²
Area 3 = (6.55)(2.25)
= 14.7375 in²
Total area = 14.7375 + 9.675 + 14.7375
= 39.15 in²
Answer:
18°
Step-by-step explanation:
A diameter cuts a circle into two arcs that are each 180°. An inscribed angle that intercepts an arc of 180° has a measure of 90°. Thus we have ...
... 5x = 90°
.. 90°/5 = x = 18°
<h3>
Answer: C) 3</h3>
The rule we'll use is a^b*a^c = a^(b+c). So we add the exponents.
That means 5^n*5^3 = 5^(n+3)
So 5^n*5^3 = 5^6 turns into 5^(n+3) = 5^6
The bases are equal to 5, so the exponents be equal to one another.
n+3 = 6
n+3-3 = 6-3
n = 3
So 5^3*5^3 = 5^(3+3) = 5^6.
That doesn’t make sense. Did you mean 15?
Answer:
Step-by-step explanation:
<u><em>The complete question is</em></u>
A cone and a triangular pyramid have a height of 9.3 m and their cross-sectional areas are equal at every level parallel to their respective bases. The radius of the base of the cone is 3 in and the other leg (not x) of the triangle base of the triangular pyramid is 3.3 in
What is the height, x, of the triangle base of the pyramid? Round to the nearest tenth
The picture of the question in the attached figure
we know that
If their cross-sectional areas are equal at every level parallel to their respective bases and the height is the same, then their volumes are equal
Equate the volume of the cone and the volume of the triangular pyramid
![\frac{1}{3}\pi r^{2}H=\frac{1}{3}[\frac{1}{2}(b)(h)H]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E%7B2%7DH%3D%5Cfrac%7B1%7D%7B3%7D%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29H%5D)
simplify
we have
substitute the given values
solve for x
