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
Step-by-step explanation:
The system of equations for eq 1 which is 3x + y = 118 represents the Green High School which filled three buses(with a specific number of students identified as x) and a van(with a specific number of students identified as y) with a total of 118 students.
for eq 2; 4x + 2y = 164; represents Belle High School which filled four buses(with a specific number of students identified as x) and two vans(with a specific number of students identified as y) with a total of 164 students.
The solution represents the specific number of students in the buses and vans in eq1 and eq 2 with x being 36 students and y being 10 students.
substituting 36 for x and 10 for y in eq 1;
3(36) + 10 = 108 + 10 = 118 total students for Green High School
substituting 36 for x and 10 for y in eq2;
4(36) + 2(10) = 144 + 20 = 164 total students for Belle High school
Answer:
The answer is perpendicular to the base.
Step-by-step explanation:
Answer:
length, width, and height are (b+2), (b-2), (b+3)
Step-by-step explanation:
Doing what the problem statement tells you to do, you get ...
(b^3 +3b^2) -(4b +12)
= b^2(b +3) -4(b +3) . . . . . factor each pair of terms
= (b^2 -4)(b +3) . . . . . . . . . write as a product
= (b -2)(b +2)(b +3) . . . . . . use the factoring of the difference of squares
The three factors are (b-2), (b+2), and (b+3). We have no clue as to how to associate those with length, width, and height. We just know these are the dimensions of the box.
Step-by-step explanation:
