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2 years ago

Q + 5/9= 1/6 from big ideas mathbook

2 answers:

6 0

9We have such eqation
Q $+ \frac{5}{9} = \frac{1}{6}$ /*9 (multiply both sides by 9)
9Q+5= $\frac{9}{6}$ /-5 both sides
9Q= $\frac{9}{6}- 5$
9Q= $\frac{9}{6}- \frac{5*6}{6}$
9Q= $\frac{9-30}{6}$
9Q=- $\frac{21}{6}$ /:9 divide both sides by 9
Q=- $\frac{21}{6}:9$
Q $\frac{21}{6*9} =- \frac{21}{54}=[tex] \frac{7}{18}$ - its the answer

babymother [125]2 years ago

3 0

Q+5/9=1/6
Q+10/18=3/18
-10/18 -10/18
Q=-7/18

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Answer:

(a)

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(b)

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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

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(b)

Since, it is in units^2

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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)

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