All categories

3 years ago

Can a pyramid have rectangles as lateral faces?

2 answers:

4 0

No, it is not possible to have rectangular lateral sides. A pyramid is formed with a polygonal base (aka a base that is a polygon) and the lateral sides are triangles. The lateral sides of any pyramid are always triangles. The triangular faces all meet at the apex point, which is the highest point on the pyramid.

Side Note: The pyramids in Egypt are an example of a rectangular pyramid. This is a pyramid with a rectangular base.

mezya [45]3 years ago

3 0

Answer is no, only the base can be a rectangle

You might be interested in

Tina painted 1/2 of a bedroom in 3/4 a day. How many days will it take dor Tina to paint one bedroom

Tpy6a [65]

Answer: 1 1/2 days because 3/4+3/4 is 1 1/2

Step-by-step explanation:

I did the math can I get brainliest

7 0

3 years ago

If x = 5, find the value of x²

mihalych1998 [28]

Answer:

Step-by-step explanation:

x² is actually the same as x×x, which in this case, 5×5 = 25

4 0

3 years ago

Read 2 more answers

Given the second order homogeneous constant coefficient equation y′′−4y′−12y=0 1) the characteristic polynomial ar2+br+c is r^2-

Vitek1552 [10]

Answer:

The initial value problem $y(x) = 4 e^{-2x} -3 e^{6 x}$

Step-by-step explanation:

Step1:-

a) Given second order homogenous constant co-efficient equation

$y^{ll} - 4y^{l}-12y=0$

Given equation in the operator form is $(D^{2} -4D-12)y=0$

Step 2:-

b) Let f(D) = $(D^{2} -4D-12)y$

Then the auxiliary equation is $(m^{2} -4m-12)=0$

Find the factors of the auxiliary equation is

$m^{2} -6m+2m-12=0$

m(m-6) + 2(m-6) =0

m+2 =0 and m-6=0

m=-2 and m=6

The roots are real and different

The general solution $y = c_{1} e^{-m_{1} x} + c_{2} e^{m_{2} x}$

the roots are $m_{1} = -2 and m_{2} = 6$

The general solution of given differential equation is

$y = c_{1} e^{-2x} + c_{2} e^{6 x}$

Step 3:-

C) Given initial conditions are y(0) =1 and y1 (0) =-26

The general solution of given differential equation is

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$ .....(1)

substitute x =0 and y(0) =1

$y(0) = c_{1} e^{0} + c_{2} e^{0}$

$1 = c_{1} + c_{2}$ .........(2)

Differentiating equation (1) with respective to 'x'

$y^l(x) = -2c_{1} e^{-2x} + 6c_{2} e^{6 x}$

substitute x= o and y1 (0) =-26

$-26 = -2c_{1} e^{0} + 6c_{2} e^{0}$

$-2c_{1} + 6c_{2} = -26$ .............(3)

solving (2) and (3) by using substitution method

$substitute c_{2} =1- c_{1}$ in equation (3)

$-2c_{1} + 6(1-c_{1}) = -26$

on simplification , we get

$-2c_{1} + 6(1)-6c_{1}) = -26$

$-8c_{1} = -32$

dividing by'8' we get $c_{1} =4$

substitute $c_{1} =4$ in equation $1 = c_{1} + c_{2}$

so $c_{2} = 1-4 = -3$

now substitute $c_{1} =4 and c_{2} =-3$ in general solution

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$

$y(x) = 4 e^{-2x} -3 e^{6 x}$

now the initial value problem

$y(x) = 4 e^{-2x} -3 e^{6 x}$

7 0

4 years ago

Ill give branliest please, please, please help

mel-nik [20]

(6,-9) bc downwards means (-) so -3-6=-9

8 0

3 years ago

Read 2 more answers

Please help! Picture is included! This is due very soon and I need help! Number 11. (10 points)

AlekseyPX

Y+y+17+111+90=360
2y+128+90=360
2y=360-218
2y=142
y=142/2
y=71
thus, the angles are 71, 90, 111, 88.

8 0

3 years ago