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aleksklad [387]

3 years ago

10

Michael and Bill are 238 miles apart, traveling towards each other. If Bill travels 20 mph and Michael travels 14 mph how long u

ntil they meet?

1 answer:

storchak [24]3 years ago

5 0

Answer:

They will meet in 7 hours.

Step-by-step explanation:

They are traveling toward each other, which means that their relative speed is the sum of their speeds.

Bill travels 20 mph and Michael travels 14 mph

This means that their relative speed is of 20 + 14 = 34 mph, that is, they get 34 miles closer each hour.

Michael and Bill are 238 miles apart, how long until they meet?

Each hour, they will be 34 miles closer. How many hours for 238 miles?

1h - 34 miles

x h - 238 miles

By rule of three

$34x = 238$

$x = \frac{238}{34} = 7$

They will meet in 7 hours.

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The value of a car bought new for $28900 decreases 15% each year. Identify the function for the value of the car. Does the funct

Eva8 [605]

Answer:

<u>V (t) = 28,900 - 4,335t</u>

<u>The function clearly represent a decay</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Value of the car = $ 28,900

Annual depreciation = 15% = 0.15

2. Identify the function for the value of the car. Does the function represent growth or decay?

Let y represent the value of the car after t years of utilization, let p the price of the car and d, the annual depreciation, therefore:

V (t) = p - (d * t * p)

Replacing with the values we know:

V (t) = 28,900 - (0.15 * t * 28.900)

V (t) = 28,900 - (0.15 * t * 28.900)

<u>V (t) = 28,900 - 4,335t</u>

<u>The function clearly represent a decay.</u>

8 0

3 years ago

Factor the polynomial completely.<br><br> 9g^2 - 30gh + 25h^2

Bezzdna [24]

Answer:

Step-by-step explanation:

It's a perfect square.

It factors into (3g - 5h)^2

or (3g - 5h)(3g - 5h)

Make sure it is correctly factored.

3g*3g = 9g^2

3g * - 5h = - 15gh

-5h*3g = - 15gh

5h*5h = 25h^2

These four terms add to 9g^2 - 15gh - 15gh + 25h^2 = 9g^2 - 30gh + 25h^2 which is exactly what you started with.

7 0

3 years ago

What is the value of :-<br><br>=》<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%20-%204%7D%7B5%20-%205%7D%20" id="TexFormu

Tresset [83]

Answer:

the value is undefined

Step-by-step explanation:

hey there,

->

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

->

$\frac{0}{0}$

0 divided by 0 is undefined

4 0

2 years ago

In a right triangle one leg measures a quarter of the other, how much do its acute angles measure? with procedure please

KATRIN_1 [288]

Answer:

The answer to your question is 14° and 76°

Step-by-step explanation:

Data

leg 1 = x

leg 2 = x/4

angle 1 = ?

angle 2 = ?

Process

To solve problem use trigonometric functions. We must use the trigonometric function that relates both legs (tangent).

tangent Ф = Opposite side / Adjacent side

-Substitution

tan Ф = (x/4) / x

-Simplification

tan Ф = x / 4x

tan Ф = 1/4

- Find Ф

Ф = tan⁻¹(1/4) = 14°

- Find the other acute angle

The sum of the internal angles in a triangles equals 180°

Ф + Ф₁ + 90° = 180

-Solve for Ф₁

Ф₁ = 180 - 90 - 14

-Result

Ф₁ = 76°

3 0

3 years ago

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

3 years ago