Help Please! Find The Area Of A Circle With D=8.2

64,000 in a fraction

earnstyle [38]

Answer:

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Step-by-step explanation:

4 0

3 years ago

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Hi I need this asap if you also explain the work on how you got answers plz and thank you

weeeeeb [17]

Answer:

Q/ Draw a line ; Ans; 

$\frac{9}{5} —>1 \frac{4}{5}$

*explain ; We put the 5 in the denominator and 5 multiply 1 + 4 so equal 9 so the choice 9/5 .

Ans; 7/3—> 2 1/3

*explain ; We put the 3 in the denominator and 3 multiply 2 + 1 so equal 7 so the choice 7/3 .

Ans; 12/10 —> 1 1/5

*explain; simple (12 and 10) ÷ 2 so equal 6/5

We put the 5 in the denominator and 5 multiply 1 + 1 so equal 6 so the choice 6/5 =12/10 .

$\frac{12 \div 2}{10 \div 2} = \frac{6}{5} = 1 \frac{1}{5}$

Q/ Compare the fractions;Ans;

$\frac{2}{3} < \frac{14}{6}$

* explain; 2/3 = 0.66 and 14/6=2.33 so 2.33 greater from 0.66 so 14/6 greater from 2/3 .

$\frac{3}{8} < \frac{8}{3}$

* explain; 3/8 = 0.375 and 8/3=2.666 so 2.666 greater from 0.375 so 8/3 greater from 3/8 .

$2 \frac{1}{6} > \frac{5}{9}$

* explain; 2 1/6 —> We put the 6 in the denominator and 6 multiply 2 + 1 so equal 13 so equal 13/6

13/6 = 2.16 and 5/9=0.55 so 2.16 greater from 0.55 so 13/6 = 2 1/6 greater from 5/9 .

Q/Add; Ans;

$\frac{7}{8} + \frac{5}{8} = \frac{7 + 5}{8} = \frac{12}{8} = \frac{3}{2}$

$\frac{1}{9} + \frac{2}{3} \\ \\ \frac{1}{9} + \frac{6}{9} = \frac{ 1+6 }{9} = \frac{7}{9}$

Q/Subtract; Ans;

$\frac{6}{7} - \frac{4}{7} = \frac{6 - 4}{7} = \frac{2}{7}$

$\frac{7}{3} - \frac{2}{9} \\ \\ \frac{21}{9} - \frac{2}{9} = \frac{21 - 2}{9} = \frac{19}{9}$

Q/ Multiply;Ans;

$\frac{1}{5} \times \frac{1}{2} = \frac{1}{10}$

$\frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$

Q/Divide;Ans;

$\frac{3}{4} \div \frac{2}{5} \\ \\ \frac{3}{4} \times \frac{5}{2} = \frac{15}{8}$

$\frac{8}{9} \div \frac{1}{3} \\ \\ \frac{8}{9} \times \frac{3}{1} = \frac{24}{9} = \frac{8}{3}$

I hope I helped you^_^

5 0

2 years ago

Evaluate the expression. 0.1×( – 0.9+ – 0.2÷ – 0.5) Write your answer as an integer or a decimal. Do not round.

matrenka [14]

Answer:

-0.05

Step-by-step explanation:

Given the expression :

0.1×( – 0.9+ – 0.2÷ – 0.5)

Evaluating the bracket :

0.1 * (-0.9 - 0.2 ÷ - 0.5)

Solving the division first

-0.2 ÷ - 0.5 = 0.4

Now we have ;

0.1 * (-0.9 + 0.4)

0.1 * - 0.5

= - 0.05

4 0

2 years ago

A student states that Figure JKLM is congruent to Figure PQRS. Determine if the student is correct or has made an error. ;D

lyudmila [28]

Answer:

Student has made an error

Step-by-step explanation:

If two figures are said to be congruent, this implies that area of both is same and both as exactly same or copy or each other.

But from the graph, it can be stated that height of figure JKLM is 4 units whereas that of other is 6 unit.

Hence explained !

4 0

2 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

3 0

3 years ago

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