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4 years ago

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The ratio of Monarch butterflies to Queen butterflies at a butterfly farm was 8:5. If there are no additional butterflies at the

farm, what is the ratio of Queen butterflies to the total number of butterflies at the farm?

1 answer:

alisha [4.7K]4 years ago

8 0

The ratio would be 5:13

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Simplify the expression.

Amanda [17]

Answer:

-1.8f - 19

Step-by-step explanation:

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3 years ago

On the Sat the lowest value anyone can score on the math section is 200 and the highest is 800. Jason is a senior retaking the S

schepotkina [342]

Answer:

10 people

Step-by-step explanation:

Parking needs to be paid for but if all friends are going together, they need to only pay once for parking.

The amount spent on just tickets is therefore:

= 243.25 - 15.75

= $227.50.

If $227.50 was for tickets alone, the number of tickets they could buy is:

= Total amount left/ Price of tickets

= 227.50 / 22.75

= 10 people

Step-by-step explanation:

6 0

3 years ago

stephanie read 72 pages on sunday and 83 pages on Monday. About how many pages did Stephanie read during the two days ?

frutty [35]

155 pages because if you add 72and83 you will get 155 because 2 plus3 is 5 and 7 plus 8 is fiveteen

4 0

3 years ago

Read 2 more answers

I am just completely lost and am unsure of how to approach this problem.

Ipatiy [6.2K]

Answer:47

Step-by-step explanation: sum of angles in a straight is 180

Aob +Box =180

6x+29+3x+124 =180

9x + 153 = 180

9x =27

x= 3

Boc =6*3 +29

Boc = 47

4 0

3 years ago

the figure below shows a circle centre of radius 10 cm the chord PQ=16cm calculate the area of the shaded region

m_a_m_a [10]

Step-by-step explanation:

OPQ originally forms a sector, the formula for sector is

$\frac{x}{360} \pi {r}^{2}$

where x is the degrees of rotation between the two radii.

We know three sides length and is trying to find an angle between the radii so we can use law of cosines which states that

$16 = \sqrt{10 {}^{2} + 10 {}^{2} - 2(100) \times \cos(o) }$

This isn't the standard formula, it's for this problem

$16 = \sqrt{200 - 200 \times \cos(o) }$

$16 = \sqrt{200 - 200 \cos(o) }$

$256 = 200 - 200 \cos(o)$

$56 = - 200 \cos(o)$

$- \frac{7}{25} = \cos(o)$

$\cos {}^{ - 1} ( - \frac{7}{25} ) = \cos {}^{ - 1} ( \cos(o) )$

$106.26 = o$

So we found our angle of rotation, which is 106.26

Now, we do the sector formula.

$\frac{106.26}{360} (100)\pi$

$\frac{10626}{360} \pi$

is the area of sector.

Now let find the area of triangle, we can use Heron formula,

The area of a triangle is

$\sqrt{s(s - a)(s - b)(s - c)}$

where s is the semi-perimeter.

To find s, add all the side lengths up, 10,10,16 and divide it by 2.

Which is

$\frac{36}{2} = 18$

$\sqrt{18(18 - 10)(18 - 10)(18 - 16)}$

$\sqrt{18(8)(8)(2)}$

$\sqrt{(36)(64)}$

$6 \times 8 = 48$

So our area of the triangle is 48. Now, to find the shaded area subtract the main area,( the sector of the circle) by the area of the triangle so we get

$\frac{10626}{360} - 48$

Which is an approximate or

44.73

6 0

2 years ago