Answer:

Explanation:
In its raw form, function notation essentially represents an equation with only one unknown variable, expressed in terms of another. Thus, f(x) = x² + 7x can be expressed as
g(x) = 2x + 3
f(g(x)) = (2x + 3)²
f(g(x)) = 4x² + 12x + 9
Hope it helps :) and let me know if you want me to elaborate.
The question is incomplete , the complete question is;
The mass of each Wt isotope is measured relative to C-12 and tabulated below. Use the mass of C-12 to convert each of the masses to amu and calculate the atomic mass of Wt.
Wt-296 = 24.6622
Answer:
The atomic mass of Wt-296 is 195.9464 amu.
Explanation:
1 amu is defined as 1 by twelfth of the carbon-12 mass.
Mass of an isotope Wt-296 = 24.6622
Mass of an isotope Wt-296 in amu = M


The atomic mass of Wt-296 is 195.9464 amu.
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
An element is <span>a part or aspect of something abstract, especially one that is essential or characteristic. I hope this helps you good luck :)</span>