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Anton [14]
3 years ago
9

A sedan will drive ten miles north, 15 miles east, 13 miles south, 15 miles west, and 22 miles north. The sedan gets 20 miles pe

r gallon, and there are five gallons of gas left in the tank. Compute the relative distance and the total distance. Then decide if there is enough fuel for the trip.
Mathematics
1 answer:
larisa [96]3 years ago
3 0
You would have enough gas to go another 25 miles
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The graph of g(x) is f(x) translated to the left 8 units and up 2 units. What is the function rule for g(x) given f(x) = x²?
NNADVOKAT [17]
Translation of a function y = h(x) to the right/left is given by y = h(x + a)  for translation to the left by 'a' units and y = h(x - a) for translation to the right by 'a' units

Translation of a function y = h(x) up/down the y-axis is given by y = h(x) + a for translation 'a' unit up and y = h(x) - a for translation 'a' unit down

So, translating f(x) = x² eight units right and two units up gives:

g(x) = (x-8)² + 2
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3 years ago
The value of x in x + 7 = 15 is_____​
madreJ [45]

Answer:

x=8

Step-by-step explanation:

x+7=15

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x=8

6 0
3 years ago
Read 2 more answers
To travel 105 miles, it takes Sue, riding a moped, 3 hours less time than it takes Doreen to travel 64 miles riding a bicycle. S
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<span>Let t=time Doreen travels; let r=rate of Doreen

                RATE * TIME = DIST.
        Sue | r+13      t-3       105
   Doreen |   r           t          64
 
Now solve the system:
(r+13)(t-3)=105
</span>r*t=64<span>
 
</span>r= \frac{64}{t} &#10;\\&#10;\\(\frac{64}{t} +13)(t-3)=105 &#10;\\&#10;\\64- \frac{192} {t}+13t-39-105=0&#10;\\&#10;\\13t- \frac{192} {t}-80=0&#10;\\&#10;\\ 13t^2-80t-192=0&#10;\\&#10;\\t_{1,2}= \frac{80\pm \sqrt{(-80)^2-4\times13(-192)} }{2\times13}= \frac{80\pm128}{26}  &#10;\\&#10;\\t=\frac{80+128}{26}  =8&#10;\\&#10;\\r=\frac{64}{t} =\frac{64}{8} =8<span>
             
Doreen: rate = 8 mph, time = 8 hours
Sue:  </span>rate = 21 mph, time = 5 hours<span>


</span>
8 0
3 years ago
Adam plans to choose a video game from a section of the store where everything is
alexira [117]

Original price of video game minus the sale price (-.75d) = the percent of original price Adam will pay (.25d)

4 0
2 years ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

7 0
3 years ago
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