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2 years ago

A car is moving at a average speed of 56 hole 3/5 km per hours . How much distance will it cover in 7 hole 1/2 hours?

Mathematics

1 answer:

vovikov84 [41]2 years ago

5 0

Answer:

3.5km

Step-by-step explanation:

not sure but this is how I'll do it tho

distance= 7 x 1/2 = 3.5 (km)

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The state of mishgan recored an all time high snowball of 355.9 inches during the winter of 1978-79 its lowest snowfall of 81.3

Rama09 [41]

Answer:

<u>274.6 inches</u> more snowfall Michigan received during the winter of 1978-79 than in 1930-31.

Step-by-step explanation:

Given:

The state of Michigan recorded an all time high snowfall of 355.9 inches during the winter of 1978-79.

Its lowest snowfall of 81.3 inches was recorded during the winter of 1930-31.

Now, to find how much more snowfall did Michigan received during the winter of 1978-79 than in 1930-31.

The snowfall in 1978-79 = 355.9 inches.

And, the snowfall in 1930-31 = 81.3 inches.

So, to get the how much more snowfall in 1978-79 than 1930-31 we subtract:

Snowfall in 1978-79 - Snowfall in 1930-31.

$355.9\ inches-81.3\ inches = 274.6\ inches.$

Therefore, 274.6 inches more snowfall Michigan received during the winter of 1978-79 than in 1930-31.

8 0

2 years ago

Darius buys a bottle of a chemical solution that contains 70% alcohol. The bottle contains 500 milliliters of solution.

Andru [333]

Answer:

350 ml

Step-by-step explanation:

70% of 500 ml = 0.7*500 ml = 350 ml

6 0

2 years ago

Read 2 more answers

3 2/5 + 9 1/10= Simplest form

Talja [164]

Change each mixed fraction to improper

3 2/5 = 15/5 + 2/5 = 17/5

9 1/10 = 90/10 + 1/10 = 91/10

Find a common denominator. Multiply 2 to both numerator and denominator of 17/5

17/5(2/2) = 34/10

34/10 + 91/10 = 125/10

125/10

Simplify. Divide

125/10 = 12.5

12.5 is your answer

hope this helps

3 0

2 years ago

Read 2 more answers

Prove that

Pani-rosa [81]

Let's start from what we know.

$(1)\qquad\sum\limits_{k=1}^n1=\underbrace{1+1+\ldots+1}_{n}=n\cdot 1=n\\\\\\
(2)\qquad\sum\limits_{k=1}^nk=1+2+3+\ldots+n=\dfrac{n(n+1)}{2}\quad\text{(arithmetic series)}\\\\\\
(3)\qquad\sum\limits_{k=1}^nk\ \textgreater \ 0\quad\implies\quad\left|\sum\limits_{k=1}^nk\right|=\sum\limits_{k=1}^nk$

Note that:

$\sum\limits_{k=1}^n(-1)^k\cdot k^2=(-1)^1\cdot1^2+(-1)^2\cdot2^2+(-1)^3\cdot3^2+\dots+(-1)^n\cdot n^2=\\\\\\=-1^2+2^2-3^2+4^2-5^2+\dots\pm n^2$

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first $S_n^+$ with only positive trems (squares of even numbers) and second $S_n^-$ with negative (squares of odd numbers). So:

$\sum\limits_{k=1}^n(-1)^k\cdot k^2=S_n^+-S_n^-$

And now the proof.

1) n is even.

In this case, both $S_n^+$ and $S_n^-$ have $\dfrac{n}{2}$ terms. For example if n=8 then:

$S_8^+=\underbrace{2^2+4^2+6^2+8^2}_{\frac{8}{2}=4}\qquad\text{(even numbers)}\\\\\\
S_8^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{8}{2}=4}\qquad\text{(odd numbers)}\\\\\\$

Generally, there will be:

$S_n^+=\sum\limits_{k=1}^\frac{n}{2}(2k)^2\\\\\\S_n^-=\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\\\\\\$

Now, calculate our sum:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=
\left|\sum\limits_{k=1}^\frac{n}{2}(2k)^2-\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\right|=\\\\\\=
\left|\sum\limits_{k=1}^\frac{n}{2}4k^2-\sum\limits_{k=1}^\frac{n}{2}\left(4k^2-4k+1\right)\right|=\\\\\\$

$=\left|4\sum\limits_{k=1}^\frac{n}{2}k^2-4\sum\limits_{k=1}^\frac{n}{2}k^2+4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|=\left|4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|\stackrel{(1),(2)}{=}\\\\\\=
\left|4\dfrac{\frac{n}{2}(\frac{n}{2}+1)}{2}-\dfrac{n}{2}\right|=\left|2\cdot\dfrac{n}{2}\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\left|n\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\\\\\\
$

$=\left|\dfrac{n^2}{2}+n-\dfrac{n}{2}\right|=\left|\dfrac{n^2}{2}+\dfrac{n}{2}\right|=\left|\dfrac{n^2+n}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\\\\\\\stackrel{(2)}{=}
\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk$

So in this case we prove, that:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk$

2) n is odd.

Here, $S_n^-$ has more terms than $S_n^+$ . For example if n=7 then:

$S_7^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{n+1}{2}=\frac{7+1}{2}=4}\\\\\\
S_7^+=\underbrace{2^2+4^4+6^2}_{\frac{n+1}{2}-1=\frac{7+1}{2}-1=3}\\\\\\$

So there is $\dfrac{n+1}{2}$ terms in $S_n^-$ , $\dfrac{n+1}{2}-1$ terms in $S_n^+$ and:

$S_n^+=\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2\\\\\\
S_n^-=\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2$

Now, we can calculate our sum:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=
\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2-\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2\right|=\\\\\\=
\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}\left(4k^2-4k+1\right)\right|=\\\\\\=
\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}4k^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\$

$=\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-4\left(\dfrac{n+1}{2}\right)^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\=
\left|-4\left(\dfrac{n+1}{2}\right)^2+4\sum\limits_{k=1}^{\frac{n+1}{2}}k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|\stackrel{(1),(2)}{=}\\\\\\
\stackrel{(1),(2)}{=}\left|-4\dfrac{n^2+2n+1}{4}+4\dfrac{\frac{n+1}{2}\left(\frac{n+1}{2}+1\right)}{2}-\dfrac{n+1}{2}\right|=\\\\\\$

$=\left|-n^2-2n-1+2\cdot\dfrac{n+1}{2}\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-2n-1+(n+1)\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-2n-1+\dfrac{(n+1)^2}{2}+n+1-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-n+\dfrac{n^2+2n+1}{2}-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-n+\dfrac{n^2}{2}+n+\dfrac{1}{2}-\dfrac{n}{2}-\dfrac{1}{2}\right|=\left|-\dfrac{n^2}{2}-\dfrac{n}{2}\right|=\left|-\dfrac{n^2+n}{2}\right|=\\\\\\$

$=\left|-\dfrac{n(n+1)}{2}\right|=|-1|\cdot\left|\dfrac{n(n+1)}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk$

We consider all possible n so we prove that:

$\forall_{n\in\mathbb{N}}\quad\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk$

7 0

2 years ago

George is 14 years old. His age is 2 years more than twice his brother, Jhon's age. How old is John?

Luda [366]

Answer:

Step-by-step explanation:

14×2=28

28+2=30

i hate riddles

8 0

2 years ago

A car is moving at a average speed of 56 hole 3/5 km per hours . How much distance will it cover in 7 hole 1/2 hours?​

A car is moving at a average speed of 56 hole 3/5 km per hours . How much distance will it cover in 7 hole 1/2 hours?