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2 years ago

A tank containing 64 gallons lost 6 1/4% of this through leakage how much is remaining is the tank

Mathematics

2 answers:

jolli1 [7]2 years ago

4 0

if we take 64 to be the 100%, how much is 6¼% off of it?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 64&100\\ x&6\frac{1}{4} \end{array}\implies \cfrac{64}{x}=\cfrac{100}{6\frac{1}{4}}\implies \cfrac{64}{x}=\cfrac{\frac{100}{1}}{\frac{25}{4}}\implies \cfrac{64}{x}=\cfrac{100}{1}\cdot \cfrac{4}{25} \\\\\\ \cfrac{64}{x}=16\implies 64=16x\implies \cfrac{64}{16}=x\implies 4=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{it had}}{64}-\stackrel{\textit{leakage}}{4}\implies \stackrel{\textit{remaining}}{60}$

lilavasa [31]2 years ago

3 0

Answer:

60 gallons of gas remained in the tank.

Step-by-step explanation:

1) 6 1/4 = 6.25

2) 64*6.25% = 4 OR 64*6 1/4 = 4

3) 64 - 4 = 60

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There is a mountain with 45 bat caves in a row. Every cave has at least 2 bats and there are 490 bats in all. Any 7 caves in a r

zhannawk [14.2K]

Answer:

Step-by-step explanation:

The first cave has 7 times more bats than the last cave. So if the 45th cave has b bats, then the first cave has 7b bats.

There are 77 bats in every row of 7 caves. So if there are 7b bats in the first cave, then there are 77−7b bats in caves 2 through 7.

Since there are also 77 bats in caves 2 through 8, that means cave #8 must have 7b bats. Repeating this logic:

#1 = 7b

#2-#7 = 77−7b

#8 = 7b

#9-#14 = 77−7b

#15 = 7b

#16-21 = 77−7b

#22 = 7b

#23-28 = 77−7b

#29 = 7b

So the first 29 caves have 5(7b) + 4(77−7b) = 308 + 7b bats.

Now we do the same thing from the other end. If cave #45 has b bats, then caves #39-#44 have 77−b bats. And since caves #38-44 have 77 bats, then cave #38 has b bats. Therefore:

#45 = b

#39-44 = 77−b

#38 = b

#32-37 = 77−b

#31 = b

So caves 31 through 45 have 3b + 2(77−b) = 154 + b bats.

Adding that to the first 29 caves, plus x number of bats in cave #30:

308 + 7b + x + 154 + b = 462 + 8b + x

We know this equals 490.

490 = 462 + 8b + x

28 = 8b + x

x is a maximum when b is a minimum, which is b = 2.

28 = 8(2) + x

x = 12

There are at most 12 bats in the 30th cave.

8 0

2 years ago

How many times larger is 2⁴⁵ than 2⁴²

andre [41]

Answer:

Step-by-step explanation:

2⁴⁵÷2⁴²=2³

2³=8

hope it helps

have a nice day

7 0

2 years ago

Read 2 more answers

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$