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Crank
2 years ago
10

Please help with this

Mathematics
1 answer:
irakobra [83]2 years ago
7 0

Answer:

1

Step-by-step explanation:

because we have to fine p *b

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The square root of 105 would lie between which two consecutive integers?
12345 [234]

Answer:

the square root of 105 lies between 10 and 11

Step-by-step explanation:

Which perfect squares "box" 105 in?  10^2 = 100 and 11^2 = 121.

The square roots of 100 and 121 are 10 and 11.  

Thus, the square root of 105 lies between 10 and 11.

8 0
3 years ago
S=n(a1+an)/2 gives the partial sum of an arithmetic sequence. What is the formula solved for an?
Veseljchak [2.6K]
S = n(a1 + an)/2
2S = n(a1 + an)
2S = na1 + nan
nan = 2S - na1
an = (2S - n a1)/n
6 0
3 years ago
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If carl was born in 1931 how old is he now
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3 years ago
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mrs.beluga is driving on a snow covered road with a drag factor of 0.2. she brakes suddenly for a deer. The tires leave a yaw ma
Aneli [31]
First, we are going to find the radius of the yaw mark. To do that we are going to use the formula: r= \frac{c^2}{8m} + \frac{m}{2}
where 
c is the length of the chord 
m is the middle ordinate 
We know from our problem that the tires leave a yaw mark with a 52 foot chord and a middle ornate of 6 feet, so c=52 and m=6. Lets replace those values in our formula:
r= \frac{52^2}{8(6)} + \frac{6}{2}
r= \frac{2704}{48} +3
r= \frac{169}{3} +3
r= \frac{178}{3}

Next, to find the minimum speed, we are going to use the formula: s= \sqrt{15fr}
where
f is <span>drag factor
</span>r is the radius 
We know form our problem that the drag factor is 0.2, so f=0.2. We also know from our previous calculation that the radius is \frac{178}{3}, so r= \frac{178}{3}. Lets replace those values in our formula:
s= \sqrt{(15)(0.2)( \frac{178}{3}) }
s= \sqrt{178}
s=13.34 mph

We can conclude that Mrs. Beluga's minimum speed before she applied the brakes was 13.34 miles per hour. 
8 0
3 years ago
What is the 10th term of the following geometric sequence?
Tom [10]
The answer is B if not sorry
5 0
3 years ago
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