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Mariulka [41]
3 years ago
12

If the range of f(x)= square root mx and the range of g(x)=m square root x are the same, which statement is true about the value

of m?
Mathematics
1 answer:
AveGali [126]3 years ago
3 0
M can be any positive real number.

Explanation:

From f(x) = √(mx) ; if x is posive m has to be positive; if x is negative m has to be negative; if x is cero m can have any value, and the range will always be 0 or positve

From g(x) = m √x; x can only be 0 or positive and the range will have the sign of m.

Given we concluded that the range of f(x) can only be  0 or positive, then me can only be 0 or positive. 
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On a recent quiz, the class mean was 70 with a standard deviation of 2.1. Calculate the z-score (to 4 decimal places) for a pers
Igoryamba
The equation to find a z score is (x- mean)/(standard deviation)

So, in this case the equation would be
(76-70)/(2.1)

For,
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castortr0y [4]
Answer: second option y = 2(x + 7/2)^2 + 1/2

Explanation:

1) given:

y = (x + 3)^2 + (x + 4)^2

2) expand the binomials:

y = x^2 + 6x + 9 + x^2 + 8x + 16

3) add like terms:

y = 2x^2 + 14x + 25

4) take common factor 2 of the first two terms:

y = 2 (x^2 + 7x) + 25

5) complete squares for x^2 + 7x

x^2 + 7x = [x +(7/2)x ]^2 - 49/4

6) substitue x^2 + 7x = (x + 7/2)^2 - 49/4 in the equation for y:

y = 2 [ (x + 7/2)^2  - 49/4] + 25

7) take -49/4 out of the square brackets.

y = 2 (x + 7/2)^2 - 49/2 + 25

8) add like terms:

y = 2(x + 7/2)^2 + 1/2

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7 0
3 years ago
Question Details
Sliva [168]

Answer:

Atomic mass unit-

It is represented by ‘amu’. It is equal to the quantity 1/12th mass of an atom of C-12.

Actual mass of one atom of C-12 = 1.9924 x 10-23 gm. = 1.9924 x 10-26 Kg.

1 amu = 1.9924 x 10 -23/12 = 1.66 x 10 -24 gm. = 1.66 x 10 -27 Kg.

Atomic mass of an element = mass of one atom of the element / 1 amu

Actual mass of an element = atomic mass of an element in amu x 1.66 x 10 -24 gm.

For ex-

Actual mass of hydrogen atom = 1.008 x 1.66 x 10 -24 gm. = 1.6736 x 10-24 gm.

Actual mass of oxygen atom = 16.00 x 1.66 x 10 -24 gm. = 2.656 x 10-23 gm.

Average Atomic weight (Mass)-

“The Average Atomic Masses of many elements are determined by multiplying the atomic mass of each isotope by its fractional abundance and adding these values and then dividing it by 100.”

Ex.- Naturally occurring carbon contains three isotopes – C12 (98.892 % abundance), C13 (1.108 % abundance), C14 (2 x 10-10 % abundance).The relative atomic masses of these isotopes are 12.000, 13.00335 and 14.00317 amu respectively.

Average atomic mass = % of I isotope x its atomic mass + % of II isotope x its atomic mass +% of III isotope x its atomic mass/100

= 12 x 98.892 + 13.00335 x 1.108 + 14.00317 x 2 x10-10 /100

Average atomic mass= 12.011 amu

Molar volume –

Molar volume of a substance is volume occupied by 1 mole of that substance .

Molar volume of solid or liquid = molar mass / density

Molar volume of ideal gas at 00C or 273 K and 1 atmosphere pressure is 22.4 litre.

Gram atomic mass(weight) or gram atom –

“Gram atomic mass(weight) of an element is the mass of Avogadro number ( 6.023 x 1023 ) of atoms of that elements in grams.”

Ex –

mass of 1 atom of oxygen = 16 amu = 16 x 1.66 x 10-24 gm

Mass of 6.023 x 1023 atoms of oxygen= 16 x 6.023 x 1023 x 1.66 x 10-24 = 16 gm.

Hence, Gram atomic weight of oxygen = 16 gm.

No. of gram atoms = mass of element in gm / atomic mass of element in gm.

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