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Ostrovityanka [42]
3 years ago
14

Find the Lowest Common Multiple of 8 and 6.

Mathematics
1 answer:
AleksandrR [38]3 years ago
4 0

Answer:

24

Step-by-step explanation:

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The polygon in the diagram is a regular octagon with center A. Find the area of the octagon to the nearest tenth of a foot. A) 2
IRISSAK [1]
Method 1: In order to find out the area of an octagon with a radius of 4 feet, we have to split the whole figure into 8 equal isosceles triangles.
Therefore, 
We will find out the area of one triangle and multiply the area with 8 to, figure out the area of the whole octagon there are 8 similar triangles and all of them will have the same area.

Method 2: From method 1, it would take time as there are too much of calculation, therefore we would go for the shortcut using the formula:

Area = 2√2 × r²

where r<span> is the radius of the octagon.
Substituting the values,
We get:
             </span>Area = 2√2 × 4²
             Area = 2√2 × 16
             Area = 2× 1.41 × 16 
             Area = 2.828 × 16 
            Area = 45.25

Rounded to the nearest tenth:

Area = 45.3 ft²


4 0
3 years ago
Question 11 please <br> 10 points <br> So yeah
MakcuM [25]

Answer:

The answer would be 2012 and 2004

Step-by-step explanation:

when you round 30.346 to the nearest tenth it is 30.4 and when you round 30.406 to the nearest tenth then it is also 30.4.

Hope it helps!

:)

6 0
3 years ago
Read 2 more answers
If f(x) = 3x² tlo, what is f(-7)?
tia_tia [17]

Answer:

Step-by-step explanation:

Be careful how you handle this.

f(-7) = 3(-7)^2

f(-7) = 3* 49     Notice the minus sign disappears. That's because there are 2 of them.

f(-7) = 147

4 0
3 years ago
How much is 3/4 of 120? And how much is 3/8 of 160?
RoseWind [281]
3/4 of 120
=3/4*120
=360/4
=90

3/8 of 160
=3/8*160
=480/8
=60
4 0
4 years ago
Read 2 more answers
ALGEBRA 2!!!!!!!!! SHOW YOUR WORK!!!!!!!!!!!!!<br> Do f(g(x)) and g(f(x))
bezimeni [28]

\bf f(x)=\cfrac{2x-3}{x+1}~\hspace{10em}g(x)=\cfrac{x+3}{2-x}&#10;\\\\[-0.35em]&#10;\rule{34em}{0.25pt}\\\\&#10;f(~~g(x)~~)\implies \cfrac{2[g(x)]-3}{[g(x)]+1}\implies \cfrac{2\left( \frac{x+3}{2-x} \right)-3}{\left( \frac{x+3}{2-x} \right)+1}\implies&#10;\cfrac{\frac{2x+6}{2-x}-3}{\frac{x+3}{2-x}+1}&#10;\\\\\\&#10;\cfrac{\frac{2x+6-6+3x}{2-x}}{\frac{x+3+2-x}{2-x}}\implies \cfrac{2x+6-6+3x}{2-x}\cdot \cfrac{2-x}{x+3+2-x}\implies \cfrac{5x}{5}\implies x


\bf \rule{34em}{0.25pt}\\\\&#10;g(~~f(x)~~)\implies \cfrac{[f(x)]+3}{2-[f(x)]}\implies \cfrac{\frac{2x-3}{x+1}+3}{2-\frac{2x-3}{x+1}}\implies \cfrac{\frac{2x-3+3x+3}{x+1}}{\frac{2x+2-(2x-3)}{x+1}}&#10;\\\\\\&#10;\cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-(2x-3)}\implies \cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-2x+3}&#10;\\\\\\&#10;\cfrac{5x}{5}\implies x


and in case you recall your inverses, when f(  g(x)  ) = x,  or g(  f(x)  ) = x, simply means, they're inverse of each other.

4 0
4 years ago
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