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3 years ago

14

Help me will give Brainlyiest

2 answers:

umka2103 [35]3 years ago

8 0

Answer: 2.25

Step-by-step explanation: The scale factor is the larger divided by the smaller. So:

18/8 = 2.25

It is clear that it is more than twice the size of the smaller ( 8 x 2 = 16) so that means that there is still 2 left over. (8/2 = 4, but since that is not the question asked, is more reasonable to say 2/8, which equals .25)

Hope this helps!

anastassius [24]3 years ago

6 0

Answer:

9/4

Step-by-step explanation:

Since the figures are dilated, the side lengths are proportional. The proportion is <em>(side of length in figure 2)/(side of similar figure length in figure 1)</em>.

That is <em>18/8</em>, which simplifies to 9/4, 9 : 4, or 9 to 4.

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3 years ago

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How do you write 19/5 as a mixed number

svetoff [14.1K]

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8 0

3 years ago

Read 2 more answers

If a central angle of measure 30° is subtended by a circular arc of length 6 meters, as is illustrated below, how many meters in

Nana76 [90]

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3 years ago

A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

goldfiish [28.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

$m\angle AOM =30^o$

In the right triangle OAM

$tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}$

so

$\frac{a}{2r}=\frac{\sqrt{3}}{3}$

we have

$r=3\ units$

substitute

$\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units$

Find the area of six equilateral triangles

$A=6[\frac{1}{2}(r)(a)]$

simplify

$A=3(r)(a)$

we have

$r=3\ units\\a=2\sqrt{3}\ units$

substitute

$A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2$

Therefore

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

7 0

4 years ago