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Nadusha1986 [10]
2 years ago
7

Marc can run 2 miles in 16 minutes at this unit rate how long will it take mars to run 5 miles

Mathematics
1 answer:
ss7ja [257]2 years ago
5 0
40 minutes to run 5 miles
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Prove Forall x elementof Z, x^2 - 5x - 1 is odd.
Diano4ka-milaya [45]

Answer with Step-by-step explanation:

Let Z be a set of integers

We have to prove that x^2-5x-1 is odd for all value of x \inZ.

There are two cases

1.When x is an odd integer number

2.when x is even integer number

1.When x is an odd integer  number

When x is positive odd integer

Then x^2  is positive odd number (because square of odd positive number is always positive an odd number )

5x is also odd number

-5x -1 = -even number

Suppose x=3

Then -15-1=-16

Odd number - even number=Odd number

Hence ,x^2-5x-1 is an odd number .

When x is negative odd number

Then  x square is positive odd number  and 5x is negative odd term term

Therefore,odd number + odd number -1=Even  number -1=Odd number

Hence x^2-5x-1 is an odd number.

2.When  x is an even number

When x is positive even number

x square is positive even number  and 5x is positive even number

Even number -Even number -1 =Even number -odd number =Odd number

Suppose x=4

x^2-5x-1=16-20-1=-5=Odd number

When x is negative even number

Then x square is positive even number  and 5x is negative even number

x^2-5x-1=Even number +Even number -1=Even number -1=Odd number

Hence, for all elements of Zx^2-5x-1is odd.

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3 years ago
Solve <br><br> x^2-4=77<br><br> E2020
Oxana [17]

Answer:

Step-by-step explanation:

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2 years ago
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Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(
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Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

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2 years ago
HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP
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Answer:

Step-by-step explanation:

Pretty sure is convergence lines

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Which quadratic function is represented by the graph?
vagabundo [1.1K]

x =(2-√24)/-2=1+√ 6 = 1.449  

x =(2+√24)/-2=1-√ 6 = -3.449

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3 years ago
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