Answer:
5 or 6
Step-by-step explanation:
So I did it in a different way than shown in rsm classes. I said that the speed was x-2 and x.
then I said the distance was 6 and 15. If the time taken was 1 hour more, then we can come up with this equation:
Since the time is 6/x-2 or 15/x
15/x-6/x-2=1
multiply everything to get a common denominator then cancel it out to get
(15x-30)-6x=x^2-2x
11x-30=x^2
x^2-11x+30=0
(x-5)(x-6) is factored form.
so x=5 or x=6. this means that the speed of the boat is either 5km/h or 6 km/h
Answer:
4.5 sq. units.
Step-by-step explanation:
The given curve is ![y = (3x)^{\frac{1}{2} }](https://tex.z-dn.net/?f=y%20%3D%20%283x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)
⇒
...... (1)
This curve passes through (0,0) point.
Now, the straight line is y = 3x - 6 ....... (2)
Now, solving (1) and (2) we get,
![y^{2} - y - 6 = 0](https://tex.z-dn.net/?f=y%5E%7B2%7D%20-%20y%20-%206%20%3D%200)
⇒ (y - 3)(y + 2) = 0
⇒ y = 3 or y = -2
We will consider y = 3.
Now, y = 3x - 6 has zero at x = 2.
Therefor, the required are = ![\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E3_0%20%7B%283x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%20-%20%5Cint%5Climits%5E3_2%20%7B%283x%20-%206%29%7D%20%5C%2C%20dx)
= ![\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5B%7B%5Cfrac%7Bx%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%5D%5E%7B3%7D%20_%7B0%7D%20-%20%5B%5Cfrac%7B3x%5E%7B2%7D%20%7D%7B2%7D%20-%206x%20%5D%5E%7B3%7D%20_%7B2%7D)
= ![[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Csqrt%7B3%7D%5Ctimes%202%20%5Ctimes%203%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B3%7D%5D%20-%20%5B13.5%20-%2018%20-%206%20%2B%2012%5D)
= 6 - 1.5
= 4.5 sq. units. (Answer)
Answer:
Part 1)
----->
Part 2)
----> ![x\neq -4](https://tex.z-dn.net/?f=x%5Cneq%20-4)
Part 3)
----> All real numbers
Part 4)
----> ![x\neq -4](https://tex.z-dn.net/?f=x%5Cneq%20-4)
Step-by-step explanation:
we know that
The domain of a function is the set of all possible values of x
Part 1) we have
![f(x)=\frac{x+4}{x}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%2B4%7D%7Bx%7D)
we know that
In a quotient the denominator cannot be equal to zero
so
For the value of x=0 the function is not defined
therefore
The domain is
![x\neq 0](https://tex.z-dn.net/?f=x%5Cneq%200)
Part 2) we have
![f(x)=\frac{x}{x+4}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%7D%7Bx%2B4%7D)
we know that
In a quotient the denominator cannot be equal to zero
so
For the value of x=-4 the function is not defined
therefore
The domain is
![x\neq -4](https://tex.z-dn.net/?f=x%5Cneq%20-4)
Part 3) we have
![f(x)=x(x+4)](https://tex.z-dn.net/?f=f%28x%29%3Dx%28x%2B4%29)
Applying the distributive property
![f*(x)=x^2+4x](https://tex.z-dn.net/?f=f%2A%28x%29%3Dx%5E2%2B4x)
This is a vertical parabola open upward
The function is defined by all the values of x
therefore
The domain is all real numbers
Part 4) we have
![f(x)=\frac{4}{x^2+8x+16}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B4%7D%7Bx%5E2%2B8x%2B16%7D)
we know that
In a quotient the denominator cannot be equal to zero
so
Equate the denominator to zero
![x^2+8x+16=0](https://tex.z-dn.net/?f=x%5E2%2B8x%2B16%3D0)
Remember that
![x^2+8x+16=(x+4)^2](https://tex.z-dn.net/?f=x%5E2%2B8x%2B16%3D%28x%2B4%29%5E2)
(![x+4)^2=0](https://tex.z-dn.net/?f=x%2B4%29%5E2%3D0)
The solution is x=-4
so
For the value of x=-4 the function is not defined
therefore
The domain is
![x\neq -4](https://tex.z-dn.net/?f=x%5Cneq%20-4)
10 cents, cause the {ball +$1.00=$1.10} subtract $1 from both sides the ball=$0.10