Answer:
A. $301
B. $721
Step-by-step explanation:
Let $x be the amount of money they raised.
Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then
Polly tried to put the $1 bills into three equal piles and found one left over at the end, then
Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then
Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then
This means 
is divisible by 2, 3, 4, 5 and 6 without remainder, so
Hence,
The smallest amount of money they could have raised is $301, because
is divisible by 7.
Now, the number 
should be divisible by 7 and must be greater than 500.
So,
When n = 9,
is not divisible by 7.
When n = 10,
is not divisible by 7.
When n = 11,
is not divisible by 7.
When n = 12,
is divisible by 7.
B. The least amount of money they could have raised is $721
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
<span>What you do now is the following: </span>
<span>Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>
<span>So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8° </span>
<span>The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram. </span>
<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>≃ 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>
<span>b) </span>
<span>volume of water = cross-sectional area * length </span>
<span>≃ 13.5 * 30 cm³ </span>
<span>≃ 404 cm³</span>
Hoped it Helped
<span>-5p^5q^4/ 8p^2q^2
= -5p^3q^2
</span><span>exponent on the p will be 3
answer
</span><span>C.3</span>
is it 46 ??? because i did the work and don't feel like explaining
Incomple question. However, here's the remaining part of the question:
14
2009
Meadow Fritillary= 5
Variegated Fritillary= 7
Zebra Swallowtail= 33
Eastern-Tailed Blue= 242
Louden County Butterfly Count
2010
Meadow Fritillary 34
Variegated Fritillary 95
Zebra Swallowtail 21
Eastern-Tailed Blue 168
2011
Meadow Fritillary
Variegated Fritillary
Zebra Swallowtail
Eastern-Tailed Blue
10
170
<u>Options</u>:
A) All butterfly populations are steadily decreasing.
B)All butterfly populations were larger than usual in 2010.
C)The Eastern-Tailed Blue butterfly is more common than the others.
D)The Meadow Fritillary is equally common as the Variegated Fritillary
Answer:
<u>C</u>
Step-by-step explanation:
Looking through the above count data by Louden County Wildlife Conservancy from 2009 to 2011 we notice the Eastern- Trailed Blue butterfly has a higher count, which implies that the Eastern-Tailed Blue butterfly is more common than the other butterflies.
Therefore, we could infer from the samples, that the Eastern-Tailed Blue butterfly is more common than others from the records of the past 3 three years.
