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3 years ago

The surface area of a sphere is 196 π . Its radius is

Mathematics

2 answers:

iren2701 [21]3 years ago

4 0

Answer:

your mom

Step-by-step explanation:

stop trying to cheat dude its not cool

Reika [66]3 years ago

4 0

Answer:

The radius of the sphere is 7 units

Step-by-step explanation:

Surface Area

Given a sphere of a radius (r), the surface area is given by the formula:

$\displaystyle A=4\pi r^2$

If the surface area is given, we can solve for r:

$\displaystyle r=\sqrt{\frac{A }{4\pi }}$

The surface area of a sphere is given A=196π.

Calculating the radius:

$\displaystyle r=\sqrt{\frac{196\pi}{4\pi }}$

$\displaystyle r=\sqrt{49}$

r = 7 units

The radius of the sphere is 7 units

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Which one of the following statements is true?

kicyunya [14]

The first derivitive is the slope
a positive 1st deriviive is positive slope or increasing
a negaitve 1st derivitive is negative slope or decreasing

the 2nd derivitive tells about the concavity
a negative second derivitive means it is concave down at that point
a positive 2nd derivitive means it is concave up at that point

so

first one is false
2nd is true
3rd is false because it could possibly be a minimum as well

answer is 2nd one

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3 years ago

Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t

Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1. $f(x)=\frac{x^2+x-20}{x^2+4}$

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345$

3. $h(x)=\frac{3x-5}{x^2-5x+7}$

$x^2-5x+7=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

$\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43$

2. $g(x)=\frac{x-17}{x^2+75}$

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12$

4. $i(x)=\frac{x^2-9}{x-9}$

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is not continuous on the set of real numbers.

5. $j(x)=\frac{4x^2-7x-65}{x^2+10}$

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0$

6. $k(x)=\frac{x+1}{x^2+x+29}$

$x^2+x+29=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

$\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102$

7. $l(x)=\frac{5x-1}{x^2-9x+8}$

$x^2-9x+8=0$

$x^2-8x-x+8=0$

$x(x-8)-1(x-8)=0$

$(x-8)(x-1)=0$

$x=8,1$

The function is not defined for x=8 and x=1

Hence, function l is not defined for all real values.

8. $m(x)=\frac{x^2+5x-24}{x^2+11}$

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722$

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

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3 years ago

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Nata [24]

Answer:

$3\frac{12}{13}$

Step-by-step explanation:

51 / 13 = 3.9231 = 3

51 - (13 x 3) = 12

$3\frac{12}{13}$

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... 0.48 × selling price = cost

... selling price = cost/0.48 = $38.87/0.48 = $80.98

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... 0.60 × selling price = $12.95

... selling price = $12.95/0.60 = $21.58

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2 years ago