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Gre4nikov [31]
3 years ago
13

Can Someone plz help? Thx

Mathematics
1 answer:
Andreyy893 years ago
8 0

Answer:

7. 2355 m

8. 1570 km

9. 157 cm

10. 40.82 m

Step-by-step explanation:

The circumference of a circle is pi * diameter or 2 * pi * radius.

If you are given the diameter, just multiply it by pi, 3.14.

If you are given the radius, then multiply the radius by 2 and then by pi, 3.14.

7. d = 750 m

circumference = pi * d = 3.14 * 750 m = 2355 m

8. d = 500 km

circumference = pi * d = 3.14 * 500 km = 1570 km

9. r = 25 cm

circumference = 2(pi)r = 2 * 3.14 * 25 cm = 157 cm

10. r = 6.5 m

circumference = 2(pi)r = 2 * 3.14 * 6.5 m = 40.82 m


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So, givens: total lesson cost of $260, total lessons taken are 6, and the first lesson costs 1.5 (or 3/2) as much as the additional lessons.

First thing to do is to figure out how many additional lessons are in that, which are 5.

Then you can make a 1 variable equation with the information you have. I’m using x as my variable.

260= 3/2x + 5x

Combine like terms.

260 = 13/2x

Divide both sides by 13/2 (treat it as a fraction, and if your calculator cannot make fractions, then decimal might help for this. 13/2=6.5)

X=40
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Lines a and b are parallel. The slope of line b is * What is the slope of line a?
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Find a positive number for which the sum of it and its reciprocal is the smallest​ (least) possible. Let x be the number and let
Allisa [31]

Answer:

S(x) = x + \frac{1}{x} --- Objective function

Interval = \{x:x=1\}

Step-by-step explanation:

Given

Represent the number with x

The required sum can be represented as:

x + \frac{1}{x}

Hence, the objective function is:

S(x) = x + \frac{1}{x}

To get the the interval, we start by differentiating w.r.t x

<em>Using first principle, this gives:</em>

S'(x) = 1 - \frac{1}{x^2}

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0 = 1 - \frac{1}{x^2}

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0 -1 = 1 -1 - \frac{1}{x^2}

-1 = - \frac{1}{x^2}

Multiply both sides by -1

1 = \frac{1}{x^2}

Cross Multiply

x^2 * 1 = 1

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Take positive square root of both sides because x is positive

\sqrt{x^2} = \sqrt{1

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Representing x using interval notation, we have

Interval = \{x:x=1\}

To get the smallest sum, we substitute 1 for x in S(x) = x + \frac{1}{x}

S(1) = 1 + \frac{1}{1}

S(1) = 1 + 1

S(1) = 2

<em>Hence, the smallest sum is 2</em>

3 0
3 years ago
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