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2 years ago

What is the slope of the line containing the points (-2, 1) and (4, -3)?

Mathematics

1 answer:

Kay [80]2 years ago

8 0

Step-by-step explanation:

the slope is always the ratio "y coordinate charge / x coordinate change" when going from one point on the line to another.

so, when going from (-2, 1) to (4, -3)

x changes by +6 (from -2 to 4).

y changes by -4 (from 1 to -3).

therefore, the slope is

-4/+6 = -4/6 = -2/3

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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

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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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