Please help so urgent
Answers
Answer:
Option E. None of the above.
Step-by-step explanation:
From the question given above, the following data were obtained:
f(x) = (x – 5)/(2x + 3)
Inverse of f(x) => f¯¹(x) =?
Recall:
When a function f(x) is multiplied by it's inverse f¯¹(x), the result is equal to 1 i.e
f(x) × f¯¹(x) = 1
With the above information, we can determine the inverse of function given above as follow:
f(x) = (x – 5)/(2x + 3)
Inverse of f(x) => f¯¹(x) =?
f(x) × f¯¹(x) = 1
(x – 5)/(2x + 3) × f¯¹(x) = 1
f¯¹(x)(x – 5) / (2x + 3) = 1
Cross multiply
f¯¹(x)(x – 5) = (2x + 3)
Divide both side by (x – 5)
f¯¹(x) = (2x + 3) / (x – 5)
Thus, the inverse of the function is (2x + 3) / (x – 5).
Option E gives the correct answer to the question.
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100% of any number must greater than or
equal to 100
Answers
For example 100% of 2 is 2
Answers
debt ratio formula can be found with
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Answers
3=c/-11
a.33
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Answers
A. y ≤ −2x + 3
y ≤ x + 3
B. y ≥ −2x + 3
y ≥ x + 3
C. y ≤ −3x + 2
y ≤ −x + 2
D. y > −2x + 3
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Answers
We can first eliminate D since it uses these (<, >) signs and the lines are shaded, indicating the points on those lines are solutions.
We can also eliminate C because the y intercept in C’s lines is 2, while in the graph, they are both 3.
Finally, we can look at both inequalities on the graph and see that the shaded areas are both underneath the line. This means that y is less than the equation for the line, eliminating B
So, the answer is A
Hope this made sense!!
Answers
solution:
a) To calculate the volume of cube of edge 1.0cm
Volume= (edge)³
=(100cm) ³
= 1cm³
To find the mass of MG atoms
Density = mass/volume
Mass of mg = 1.74g/cm³ x 1cm³
= 1.74g
To find the number of mg atoms,
Molar mass of mg = 24.31g
24.31g mg contain 6.022 x 10²³ atoms of mg
1.74g mg contain = 6.022 x 10²³mg atoms/24.31g mg x 1.74g mg
= 0.4310 x 10²³mg atoms
= 4.31 x 10²² mg atoms
B) to find the volume occupied by mg atoms
Total volume of cube = 1 cm³
Volume occupied by mg atoms
= 74% of 1cm³
=74/100 x 1cm³= 0.74cm³
To find the volume of 1 mg atom
4.31 x 10²² mg atoms occupy 0.74 cm³
1 atom will occupy 0.74cm³
1 atom will occupy = 0.74cm³/4.31 x 10²²mg atoms x 1mg atom
volume of 1 mg atom = 0.1716 x 10⁻²²cm²
Volume of 1 mg atom = 4/3 π³ in the form of sphere.
4/3πr³ = 0.1716 x 10⁻²²cm³
R³ = 0.1716 x 10⁻²²cm³x 4/3 x 7/22
R³ = 0.04095 x 10⁻²²cm³
R = 0.1599 x 10⁻⁷cm
1 = 10⁻¹⁰cm
= 0.1599 x 10⁻⁷cm
= 1pm/1 x 10⁻¹⁰cm x 0.1599 x 10-7cm
= 1.599 x 10² pm
R= 1.6 x 10²pm