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olganol [36]
4 years ago
10

Julissa is running a 10-kilometer race at a constant pace. After running for 18 minutes, she completes 2 kilometers. After runni

ng for 54 minutes, she completes 6 kilometers. Her trainer writes an equation letting t, the time in minutes, represent the independent variable and k, the number of kilometers, represent the dependent variable.
Which equation can be used to represent k, the number of kilometers Julissa runs in t minutes?
Mathematics
1 answer:
IRINA_888 [86]4 years ago
3 0
So18 is to 2 and54 is to 6
18:2 and 54:6
18/2 and 54/6
simplify
9/1 and 9/1
she is running at the same rate

so we need to do
t times something=k
its eems that she divided the time by 9
so
1/9t=k

the equation is 1/9t=k

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irina [24]

Answer:

The solution is (-5, -2).

Step-by-step explanation:

We'll use "elimination by addition and subtraction."

Multiply the second row by -2.  Our system

– 7x – 10y = 45

-3x - 5y = 25

becomes:

– 7x – 10y = 45

+6x  + 10 y = -50

-------------------------

  -x             =  5, so that x = -5.

Substituting -5 for x in the second equation results in

-3(-5) - 5y = 25, or  15 - 5y = 25.  This reduces to

-5y = 10, so that y must be -2.

The solution is (-5, -2).

4 0
3 years ago
Given P(A)= 0.95 and P(A∩B)=0.37. Find P(B∣A)
Dmitriy789 [7]

Answer:

P(A)= 0.95  \: and  \: P(A∩B)=0.37. \\  Find  \: P(B∣A) \\ P(B∣A) =  \frac{P(A∩B)}{P(A)}  \\  =  \frac{0.37}{0.95}  \\  \therefore \: P(B∣A) = 0.389

6 0
3 years ago
Que numero elevado al cubo me da 19648
Lady bird [3.3K]

Answer:

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Step-by-step explanation:

8 0
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Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

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Answer:

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Step-by-step explanation:

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9 + 12x = 15

x = 6/12 = 0.5

7 0
3 years ago
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