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3 years ago

If AB = 3, AD = 5, and DE = 6, what is the length of BC? BC II DE

Mathematics

2 answers:

natali 33 [55]3 years ago

8 0

Answer: I think its 4 if not then sorry

Step-by-step explanation:

I HOPE THIS HELPS.

never [62]3 years ago

7 0

Answer:

Step-by-step explanation:

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The correct answer is -5n+1

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How many solutions does the following equation have? 3(2+x)=3x+2=

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3 years ago

The length of the base edge of a pyramid with a regular hexagon base is represented as x. The height of the pyramid is 3 times l

sergij07 [2.7K]

Answer:

(a)

h=3x

(b)

$A=\frac{\sqrt{3} }{4} x^2$

(c)

$A=\frac{3\sqrt{3} }{2} x^2$

(d)

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

Step-by-step explanation:

We are given a regular hexagon pyramid

Since, it is regular hexagon

so, value of edge of all sides must be same

The length of the base edge of a pyramid with a regular hexagon base is represented as x

so, edge of base =x

b=x

Let's assume each blank spaces as a , b , c, d

we will find value for each spaces

(a)

The height of the pyramid is 3 times longer than the base edge

so, height =3*edge of base

height=3x

h=3x

(b)

Since, it is in units^2

so, it is given to find area

we know that

area of equilateral triangle is

$=\frac{\sqrt{3} }{4} b^2$

h=3x

b=x

now, we can plug values

$A=\frac{\sqrt{3} }{4} x^2$

(c)

we know that

there are six such triangles in the base of hexagon

So,

Area of base of hexagon = 6* (area of triangle)

Area of base of hexagon is

$=6\times \frac{\sqrt{3} }{4} x^2$

$=\frac{3\sqrt{3} }{2} x^2$

(d)

Volume=(1/3)* (Area of hexagon)*(height of pyramid)

now, we can plug values

Volume is

$=\frac{1}{3}\times\frac{3\sqrt{3} }{2} x^2\times (3x)$

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

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