Answer:
The first bottle holds 25 3/5 ounces of water.
Step-by-step explanation:
16 x 8/5
= 16/1 x 8/5
= 128/5
= 25 3/5
Answer:
estimated gpa: 2.85
Step-by-step explanation:
Evaluate y=0.14x+0.75 at x = 15 (hours):
y = 0.14(15 hours) + 0.75
= 2.1 + 0.75
y = 2.85 = estimated gpa corresponding to 15 houirs of study
Answer:
the answer is (C) 2(1-2x)=x-3
Answer:
4a + 4b - 2
Step-by-step explanation:
To evaluate h(a + b) , substitute x = a + b into h(x), that is
h(a + b) = 4(a + b) - 2 = 4a + 4b - 2
Given:
Consider the given function:
![f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7Bb%5Ccdot%28x-a%29%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%5Ccdot%28x-b%29%7D%7Ba-b%7D)
To prove:
![f(a)+f(b)=f(a+b)](https://tex.z-dn.net/?f=f%28a%29%2Bf%28b%29%3Df%28a%2Bb%29)
Solution:
We have,
![f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7Bb%5Ccdot%28x-a%29%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%5Ccdot%20%28x-b%29%7D%7Ba-b%7D)
Substituting
, we get
![f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}](https://tex.z-dn.net/?f=f%28a%29%3D%5Cdfrac%7Bb%5Ccdot%28a-a%29%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%5Ccdot%20%28a-b%29%7D%7Ba-b%7D)
![f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}](https://tex.z-dn.net/?f=f%28a%29%3D%5Cdfrac%7Bb%5Ccdot%200%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%7D%7B1%7D)
![f(a)=0+a](https://tex.z-dn.net/?f=f%28a%29%3D0%2Ba)
![f(a)=a](https://tex.z-dn.net/?f=f%28a%29%3Da)
Substituting
, we get
![f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}](https://tex.z-dn.net/?f=f%28b%29%3D%5Cdfrac%7Bb%5Ccdot%28b-a%29%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%5Ccdot%20%28b-b%29%7D%7Ba-b%7D)
![f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}](https://tex.z-dn.net/?f=f%28b%29%3D%5Cdfrac%7Bb%7D%7B1%7D%2B%5Cdfrac%7Ba%5Ccdot%200%7D%7Ba-b%7D)
![f(b)=b+0](https://tex.z-dn.net/?f=f%28b%29%3Db%2B0)
![f(b)=b](https://tex.z-dn.net/?f=f%28b%29%3Db)
Substituting
, we get
![f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}](https://tex.z-dn.net/?f=f%28a%2Bb%29%3D%5Cdfrac%7Bb%5Ccdot%28a%2Bb-a%29%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%5Ccdot%20%28a%2Bb-b%29%7D%7Ba-b%7D)
![f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}](https://tex.z-dn.net/?f=f%28a%2Bb%29%3D%5Cdfrac%7Bb%5Ccdot%20%28b%29%7D%7Bb-a%7D%2B%5Cdfrac%7Ba%5Ccdot%20%28a%29%7D%7B-%28b-a%29%7D)
![f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}](https://tex.z-dn.net/?f=f%28a%2Bb%29%3D%5Cdfrac%7Bb%5E2%7D%7Bb-a%7D-%5Cdfrac%7Ba%5E2%7D%7Bb-a%7D)
![f(a+b)=\dfrac{b^2-a^2}{b-a}](https://tex.z-dn.net/?f=f%28a%2Bb%29%3D%5Cdfrac%7Bb%5E2-a%5E2%7D%7Bb-a%7D)
Using the algebraic formula, we get
![[\because b^2-a^2=(b-a)(b+a)]](https://tex.z-dn.net/?f=%5B%5Cbecause%20b%5E2-a%5E2%3D%28b-a%29%28b%2Ba%29%5D)
![f(a+b)=b+a](https://tex.z-dn.net/?f=f%28a%2Bb%29%3Db%2Ba)
[Commutative property of addition]
Now,
![LHS=f(a)+f(b)](https://tex.z-dn.net/?f=LHS%3Df%28a%29%2Bf%28b%29)
![LHS=a+b](https://tex.z-dn.net/?f=LHS%3Da%2Bb)
![LHS=f(a+b)](https://tex.z-dn.net/?f=LHS%3Df%28a%2Bb%29)
![LHS=RHS](https://tex.z-dn.net/?f=LHS%3DRHS)
Hence proved.