3 years ago

Darius takes his family on an afternoon boat ride.

1 answer:

6 0

The statement that is true would be later on as he crosses Stover Lake, he drives 30 minutes at the same average speed. Hope this helps

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A praying mantis is an interesting insect that can rotate its head 180 degrees.Suppose the praying mantis at the right is 10.5 c

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10.5 = 10 1/2 explanation is 1÷2 =0.5

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3 years ago

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A sandwich shop serves 4 ounces of meat and 3 ounces of cheese on each sandwich. After making sandwiches for an hour, the shop o

mote1985 [20]

Answer:

7 oz, 13 sandwiches, 22.75 oz, and 30.33 oz.

Step-by-step explanation:

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3 years ago

Write the slope intercept form of the equation of the line through the given point with the given slope through (-2,5),slope-3/2

hjlf

Answer: y= -3/2x + 2
y= -3/2 + b
5 = -3/2 * -2 + b
5 = 3 + b
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3 years ago

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

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Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

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3 years ago

A particle moves along the x-axis with velocity v(t) = t2 - 1, with t measured in seconds and v(t) measured in feet per second.

Kisachek [45]

Answer:

0.667 feet

Step-by-step explanation:

To get the total distance travelled, we would integrate the velocity v(t) equation with respect to time t.

Given that the particle moves from t = 0 to t = 2 seconds

Integrating v(t) = t2 - 1

we get

= t^3/3 - t

substituting the boundaries from t = 0 to t = 2 seconds, we get

= 2^3/3 - 2 - (0^3/3 - 0)

= 8/3 - 2 - 0

= 0.667 feet

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3 years ago

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