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Sloan [31]
3 years ago
7

83,400000 millimeters in kilometers

Mathematics
2 answers:
Rom4ik [11]3 years ago
7 0

Answer:

search bar on google and safari

Step-by-step explanation:

nikitadnepr [17]3 years ago
3 0

Hey there!

83,400000 millimeters = 83.4 kilometers

Hope this helps you!

God bless ❤️

xXxGolferGirlxXx

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I need help with 14, 15, and 16 please!!
stellarik [79]

Answer:

so 80 plus 20 is a 100

Step-by-step explanation:

thats it

7 0
3 years ago
What is the perimeter of the triangle below? Round to the nearest hundredth.
hoa [83]
I think - 14 or 14 but I am not sure.
3 0
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)

6 0
3 years ago
Amanda needs to make a garden plot which has an area less than 18 sq. feet. The length should be 3 feet longer than the width. W
SCORPION-xisa [38]

Answer:

<em>Any width less than 3 feet</em>

Step-by-step explanation:

<u>Inequalities</u>

The garden plot will have an area of less than 18 square feet. If L is the length of the garden plot and W is the width, the area is calculated by:

A = L.W

The first condition can be written as follows:

LW < 18

The length should be 3 feet longer than the width, thus:

L = W + 3

Substituting in the inequality:

(W + 3)W < 18

Operating and rearranging:

W^2 + 3W - 18 < 0

Factoring:

(W-3)(W+6)<0

Since W must be positive, the only restriction comes from:

W - 3 < 0

Or, equivalently:

W < 3

Since:

L = W + 3

W = L - 3

This means:

L - 3 < 3

L < 6

The width should be less than 3 feet and therefore the length will be less than 6 feet.

If the measures are whole numbers, the possible dimensions of the garden plot are:

W = 1 ft, L = 4 ft

W = 2 ft, L = 5 ft

Another solution would be (for non-integer numbers):

W = 2.5 ft, L = 5.5 ft

There are infinitely many possible combinations for W and L as real numbers.

6 0
3 years ago
What is the surface area of the cube below?
balu736 [363]

Answer: A=6a2

Step-by-step explanation:

did it help?^﹏^

6 0
3 years ago
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