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3 years ago

A human body has about 6x10^2 times as many red blood cells

Mathematics

2 answers:

algol [13]3 years ago

8 0

10x10=100

6x100=600

A human body has about 600 times as many red blood cells.

hope it helps!

Setler [38]3 years ago

7 0

Answer:

Humans are complex organisms made up of trillions of cells, each with their own structure and function.

Scientists have come a long way in estimating the number of cells in the average human body. Most recent estimates put the number of cells at around 30 trillion. Written out, that’s 30,000,000,000,000!

These cells all work in harmony to carry out all the basic functions necessary for humans to survive. But it’s not just human cells inside your body. Scientists estimate that the number of bacterial cells in the human body likely exceeds the number of human cells.

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Find the value of 1 and 1/10-1/5

Andreyy89

9514 1404 393

Answer:

9/10

Step-by-step explanation:

(1 1/10) - 1/5 = (1 1/10) -2/10 = 1 + (1/10 -2/10) = 1 -1/10 = 9/10

1.1 -0.2 = 0.9 = 9/10

5 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

Factorize 16x - 6 and show correct and full working.

Vlad1618 [11]

\left[x \right] = \left[ \frac{3}{8}\right][x]=[83] totally answer

8 0

2 years ago

Como dividir 8.2 por 2.87

lakkis [162]

It would equal 11.07. it is right

3 0

3 years ago

Read 2 more answers

Simplify x^2-9x-22/x-11

OleMash [197]

Answer:

x^2-11/-8x

Step-by-step explanation:

We can solve this equation easily by using PEMDAS, or order of operations.

Parenthesis

Exponents

Multiplication>Division (Depends on which comes first left to right)

Addition>Subtraction (See multiplication>division)

Now that we know the order of the operations, we can solve.

First, we cant solve for x^2 because we dont know the value of x, so well leave that alone. Same with 22/x. All we can do is use the commutative property.

So our equation is now x^2-11/-8x

Hope this helps!

6 0

3 years ago