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seraphim [82]
3 years ago
5

A circle has a radius of 50 cm. Which of these is the closest to its area?

Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

  7,854 cm squared

Step-by-step explanation:

The formula for the area of a circle is ...

  A = πr^2

Filling in the given radius gives you ...

  A = π(50 cm)^2 = 2500π cm^2 ≈ 7854 cm^2

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Find the slope of the line that passes through G(−3, 2) and H(7, 2).
Svetllana [295]

Answer:

y=0x+2 meaning the slope is equal to 0

Step-by-step explanation:

If you think about this question you'll see that our two points have the same y coordinate

This means that the slope is equal to 0

so we have

y=0x+b  

If we plug in -3 for x and set the equation equal to 2 we can find b

0+b=2

b=2

so we have

0x+b=y

Hope this helps!

7 0
2 years ago
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5) Choose the inequality that will help solve the following situation: A jet ski rental charges a flat rate of $75, plus an addi
yanalaym [24]

Answer:

22.5

Step-by-step explanation:

You time 75 by 60. Then, you divided it by 200. The answer is 22.5.

Have A Good Day :)

7 0
3 years ago
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What is the problem of this solving?!
nika2105 [10]
     This question can be solved primarily by L'Hospital Rule and the Product Rule.

y= \lim_{x \to 0}  \frac{x^2cos(x)-sin^2(x)}{x^4}
 
     I) Product Rule and L'Hospital Rule:

y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}
 
     II) Product Rule and L'Hospital Rule:

y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}
 
     III) Product Rule and L'Hospital Rule:

]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\  \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}
 
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y = \phi + \varphi \\  \\ \phi = \lim_{x \to 0}  \frac{-6cos(x)-[-4xsin(x)+4cos(x)]-[2cos(x)-2xsin(x)]}{24x}  \\ \varphi = \lim_{x \to 0}  \frac{[2xsin(x)+x^2cos(x)]+[8cos^2(x)-8sin(x)]}{24x}
 
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3 0
2 years ago
ASAP On the coordinate plane below, quadrilaterals TRAP and HELP are similar to each other.
ipn [44]

Answer:

1 : 2

Step-by-step explanation:

Since the quadrilaterals are similar then the ratios of corresponding sides are equal, that is

\frac{HP}{TP} = \frac{2}{4} = \frac{1}{2}

Thus the ratio of perimeters is also 1 : 2

5 0
3 years ago
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Equation of a line with a slope of -3 that passes through point (6,9)?
Serjik [45]

Answer:

<u>Point-slope form: y-9 = -3(x - 6)</u>

<u>Standard form: y = -3x+27</u>

Step-by-step explanation:

The point-slope form equation is y - y1 = m(x - x1). Because we are trying to find the equation that passes through (6,9), use these values as x1 and y1. In this equation m is the slope and we are also given that the slope is -3. This means that the equation in point-slope form is y-9 = -3(x - 6).

If it is asked for the equation in standard form, some simple moving of terms can accomplish that. Multiply out the -3 and get y-9 = -3x + 18. Add 9 to both sides to get a final answer of y = -3x+27.

7 0
3 years ago
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