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Vladimir79 [104]
3 years ago
7

Tom know that in his school 10 out of every 85 students are left handed.There are 493 students in toms school.How many students

in toms school are left handed?
Mathematics
1 answer:
sergejj [24]3 years ago
7 0

There are <u>58 left-handed students.</u>

<h3>EXPLANATION</h3>

493 ÷ 85 = 5.8

5.8 x 10 = <u>58 students</u>

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Christopher weighs 211 pounds and has a BMI of 28.5. How many whole ponds must be gain or lose to get into the healthy BMI range
eduard
BMI = (pounds * 703) / (height^2)

so in the information you are missing the height
let's find it :)
first plug in what you know:
28.5 = (211 * 703) / (height)^2   then solve for (height)
28.5 (height)^2 = (211 * 703)     multiplied both sides by (height)^2
(height)^2 = (211 * 703) / 28.5   divided both sides by 28.5
(height) = square root [(211 * 703) / 28.5]  square rooted both sides
(height) = 72.143 inches

The healthy BMI is about 25
so 25 = (pounds)(703) / (72.143^2)
25(72.143^2) / (703) = (pounds)  divide by 703, multiply by (72.143^2)
pounds = 185

211 - 185 = 26
Therefore Christopher would have to lose 26 pounds.

*Note: If the BMI you wanted to use was not 25, then just replace it like this
(pounds) = (BMI)(72.143^2) / (703) = goal weight
5 0
3 years ago
Please help with 7, 8, and 9, please put in step by step as this is a study guide for a test!
Anika [276]

Answer:

  7. f(x) = -6/7x +12/7

  8a. (-∞, -3)∪(-3, ∞); b. f(6) = 2/3

  9a. f(-3) = -33; b. f(5a) = -50a² +25a

Step-by-step explanation:

7. You are given two points: (2, 0), (-5, 6). It is often useful to start with the 2-point form of the equation of a line:

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  y = (6 -0)/(-5-2)(x -2) +0

  y = -6/7(x -2) . . . simplify

  y = -6/7x +12/7 . . . slope-intercept form

  f(x) = -6/7x +12/7 . . . . the desired functional form

__

8.

a. The <em>domain</em> is the set of x-values for which the function is defined. It will be undefined when the denominator is zero. So, the domain is all real numbers except x=-3. That can be written as ...

  -∞ < x < -3 ∪ -3 < x < ∞

b. Put 6 where x is and do the arithmetic.

  f(6)=\dfrac{6}{6+3}=\dfrac{6}{9}\\\\\boxed{f(6)=\dfrac{2}{3}}

__

9.

a. Put -3 where x is and do the arithmetic.

  f(-3)=-2(-3)^2+5(-3)=-2\cdot9-15\\\\\boxed{f(-3)=-33}

b. Put 5a where x is and simplify.

  f(5a)=-2(5a)^2+5(5a)=-2(25a^2)+25a\\\\\boxed{f(5a)=-50a^2+25a}

6 0
3 years ago
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