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Veronika [31]
3 years ago
5

A study showed that 7/10 of the human body weight is water. If a person weighs 140 pounds, how many pounds is water?

Mathematics
2 answers:
denis23 [38]3 years ago
7 0

\huge \boxed{\mathfrak{Question} \downarrow}

  • A study showed that 7/10 of the human body weight is water. If a person weighs 140 pounds, how many pounds is water?

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

Okay, so first let's note down the points we've in the question.

  • Fraction of body weight that's water = <u>7</u><u>/</u><u>1</u><u>0</u>
  • Weight of the person = <u>1</u><u>4</u><u>0</u><u> </u><u>pounds.</u>
  • Number of pounds that's water = ?

To find how many pounds of the person who weighs 140 pounds is water, we need to multiply 140 into 7/10.

⇨ Let's solve it now.

\sf \: 140 \times  \frac{7}{10}  \\ \sf  =  14 \bcancel{0} \times  \frac{7}{1 \bcancel{0}}  \\  \sf = 14 \times 7 \\  =    \boxed{\boxed{\bf \: 98 \: pounds}}

  • A person who weighs 140 pounds will be approximately <u>9</u><u>8</u><u> </u><u>pounds</u><u> </u><u>water.</u>

dmitriy555 [2]3 years ago
3 0

Answer:

98

Step-by-step explanation:

it would be 70 percent of 140

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Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$           $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$         (with error 0.0001)

 

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

3 0
3 years ago
What is the value of -17 - (-1) + 2/3 + -1/2?
netineya [11]

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Step-by-step explanation:

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3 years ago
Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point
Maksim231197 [3]

Answer: (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

or  (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]

Taking square on both the sides , we get

(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0

Using quadratic formula : x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}

so, (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

since \sqrt{3}=1.732

so, (6+7(1.732),0)\text{ and }(6-7(1.732),0) are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

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Answer:

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Step-by-step explanation:

hope this helps! :D

have a miraculous day!! <3

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5 0
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