Please summit your question by clicking the comment box. We will answer your question as soon as possible.
Saturday, January 26, 2008
Subscribe to:
Post Comments (Atom)
Instant Online Tutoring
Search Financial Aid and Schloarship Online
Educate yourself online:
6 comments:
Challenging Algebra Question?
A ring is made out of gold and copper. Gold has a density of 19.3 g/cm3. Copper has a density of 9 g/cm3. Mass m, density d, and volume v are related by the formula m=dv. The ring has a volume of 4.2 cm3, and a mass of 52.22 g.
Let a = volume of gold. Mass of gold = dv = 19.3a
Let c = volume of copper. Mass of copper = dv=9c
A. Solve the following system by elimination to find out how many grams of gold are in the ring.
a +c = 4.2
19.3a + 9c = 52.22
B. What is the percent of gold by mass?
Let a + c = 4.2 be Equation #1
Let 19.3a + 9c = 52.22 be Equation #2
Equation multiply by -9 becomes -9a - 9c = 37.8 become Equation #3.
Now, Equation #2 minus Equation #3 becomes
10.3a = 14.42 >>>>> a = 1.4 cm
2: Mass of Gold = 19.3a = 19.3 * 1.4 = 27.02 grams
Mass percentage of Gold = (mass of gold / total mass) * 100
= (27/52.22) * 100 = 51.74%
Find the domain of the function:
h(s) = sqrt(s-1)/s-4
Determine whether or not y is a function of x and state your reasons:
x^2+y^2-2x-4y+1=0
Write the given equation of the circle in standard form:
3x^2+3y^-6y-1=0
Please help me:( I'm so stressed, and I'm just not understanding this.
Problem #1:
The domain of h(s) = sqrt(s-1)/s-4
Part A (Numerator):
Because we have square root here in the numerator, s - 1 must be greater than zero. Therefore we have:
s-1>=0 and solving for s, we have
s>=1
Part B (Denominator):
The denominator in this equation can't be equal to zero. Therefore we have s-4 not equal to 0, and s is not equal to 4.
Finally, combining part A and part B, the domain of the function is:
[1,4) U (4, infinity). In English, this means s is equal to or greater than one, but d id not equal to four.
Problem #2:
This equation is a circle. To determine the equation of the circle, you must do the following by completing the square.
x^2 + y^2 - 2x - 4y + 1 = 0 becomes
(x^2-2x+1)+(y^2-4y+4)+ (1-1-4)=0
Notice above I have add one in the first (), add 4 in the second (), then subtract them again in the third (). Now the equation becomes:
(x-1)^2 + (y-2)^2 = (-1+1+4)
Finally we have:
(x-1)^2 + (y-2)^2 = 4 = 2^2
The form of the cicle is:
(x-h)^2 + (y-k)^2 = r^2
The center is at (1,2) and the radius is sqrt(4) or 2.
Since this is an equation of a circle, by the vertical line test, this equation is not a function.
Problem #3:
There's a problem with your question, please make sure you have submitted the correct one.
I assume that you mean:
3x^2 + 3y^2 - 6y - 1 = 0
Rearranging the equation, we have:
3x^2 + 3y^2 - 6y = 1
Factoring out the left hand side:
3(x^2) + 3(y^2 - 2y) = 1
Divide both side by 3:
x^2 + (y^2 - 2y) = 1/3
complete the square:
x^2 + (y^2 - 2y + 1) = 1/3 + 1
x^2 + (y-1)^2 = 4/3
So the center is at (0,1) and the radius is sqrt(4/3) or
2/sqrt(3) = 2*sqrt(3)/3
Does this help? If it does, please submit your comment and let others know by spreading the words out. Thanks!
If you have f(x)=3 square root 4a how do you find domain and range.
find the domain of this square root function.
y= square root of 2x-5
Post a Comment