(1) y varies partly as D an partly as the cube root of D.when Y=6, D=27 an when Y=11,D=1/8.find Y when D=3. Correct to 2 decimal point.
(2) using completing the square method find the root of the equation
8x+6x-5=0
1) y = aD + b (cube root of D)
a and b are constant.
when y=6, d=27
6= 27a + b(3√ 27)
6= 27a + 3b ----------equation 1
when y= 11, d = 1/8
11= a/8 + b( 3√ 1/8)
11= a/8 + b/2 --------- equation 2
so we have
6= 27a + 3b
11= a/8 + b/2
Rearrange
27a + 3b =6
a/8 + b/2 =11
Multiply equation 2 by 6 and equation 1 by 1
27a + 3b =6
6a/8 + 6b/2 =66
therefore,
27a + 3b =6
3a/4 + 3b =66
Eliminate 3b in both equation
27a =6
3a/4=66
subtract equ 1 from 2
-105a/4 = 60
Cross Multiply
-105a=240
a= -2.28
Solve for b
6= 27a + 3b
6= 27(-2.28) + 3b
3b=6 - (-61.56)
3b=67.56
b=22.52
Now solve for y when D=3
y = aD + b (cube root of D)
y = (-2.28 x 3) + 22.52 (cube root of 3)
y= -6.84 + 32.48
y= 25.64 to 2 D.P
Y=k(D)+j(D^1/3)
6=27k + 3j
11=k/8+j/2
multiply second equation by 6, then subtract first equation from second.
6=27k+3j
66=3k/4+5j
or
60=-k(3/4-27)
solve for k
then put that k into either equation, solve for j.
Now solve for y when D=3
(1) Ah, the mathematics of y and D! It seems like y is playing a little game with D and the cube root of D. Let's figure it out together!
When y = 6 and D = 27, y varies partly as D. So we have y = kD, where k is some constant. Plugging in the values we have: 6 = 27k. Solving for k, k = 6/27 = 2/9.
Now, when y = 11 and D = 1/8, y varies partly as the cube root of D. So we have y = h∛D, where h is some other constant. Substituting in the values we have: 11 = h∛(1/8). Rearranging and solving for h, h = 11 / (1/2) = 22.
Now, we need to find y when D = 3. Let's use the information we've gathered so far. Since y varies partly as D, we know y = kD. Plugging in the values, y = (2/9) * 3 = 2/3 * 3 = 2. The answer is y = 2.
So, when D = 3, y is a great performer and takes on the value of 2. Bravo, y!
(2) Ah, completing the square method! It's like giving the quadratic equation a little makeover. Let's see what we can do with 8x + 6x - 5 = 0.
First, let's gather all the x terms together and move the constant to the other side: 14x = 5.
Next, let's divide both sides by 14 to isolate x: x = 5 / 14.
And voila! We have the root of the equation. x = 5/14. Keep in mind that we haven't fully simplified it, but we've completed the square and found the value of x.
So there you have it, x = 5/14. Hoping your quadratic equation is now feeling fabulous and ready to take on the world!
(1) To solve for Y when D=3 using the given information, let's break down the steps:
Step 1: Determine the relationship between Y and D. From the information provided, Y varies partly as D and partly as the cube root of D. This can be written as a proportional relationship:
Y = k * D * (D^(1/3))
where k is a constant.
Step 2: Use the given data points to find the value of k.
When Y=6 and D=27:
6 = k * 27 * (27^(1/3))
Simplifying the equation:
6 = k * 27 * 3
Solving for k:
k = 6 / (27 * 3)
k = 2 / 27
Step 3: Substitute the value of k into the proportional relationship equation:
Y = (2/27) * D * (D^(1/3))
Now, we need to find Y when D=3:
Y = (2/27) * 3 * (3^(1/3))
Y = (2/27) * 3 * 1.442
Y ≈ 0.296
Therefore, when D=3, Y is approximately 0.296.
(2) To use the completing the square method to find the roots of the equation 8x + 6x - 5 = 0, follow these steps:
Step 1: Reorganize the equation by combining like terms:
14x - 5 = 0
Step 2: Move the constant term to the other side:
14x = 5
Step 3: Divide the equation by the coefficient of x:
x = 5/14
Therefore, the root of the equation 8x + 6x - 5 = 0 is x = 5/14.
To find the value of Y when D is equal to 3 in the first problem, we need to first determine the relationship between Y and D.
The problem states that Y varies partly as D and partly as the cube root of D. Let's break down the problem statement:
(1) y varies partly as D: This means that there is a direct relationship between y and D. We can write this as y = kD, where k is a constant.
(2) y varies partly as the cube root of D: This means that there is an indirect relationship between y and D. We can write this as y = k∛D, where k is a constant.
Combining these two relationships, we have y = kD * k∛D.
Now let's use the given values to find k. When Y = 6, and D = 27:
6 = k * 27 * k∛27.
Simplifying this equation, we have:
6 = 27k^2∛3.
Now, solve for k:
k^2∛3 = 6/27.
k^2∛3 = 2/9.
Taking the cube of both sides, we have:
k^2 = (2/9)^3.
k^2 = 8/729.
Taking the square root of both sides, we get:
k = √(8/729).
k = 2/27.
Now we have the value of k. We can substitute this back into our equation:
y = (2/27) * D * (∛D).
To find Y when D = 3, substitute D = 3 into the equation:
Y = (2/27) * 3 * (∛3).
Y = 2/9 * (∛3).
Calculating the cube root of 3, we have:
Y ≈ 2/9 * 1.442.
Y ≈ 0.32 (rounded to 2 decimal places).
So when D = 3, Y ≈ 0.32.
Moving on to the second problem:
To solve the equation 8x + 6x - 5 = 0 using the completing the square method, follow these steps:
1. Rearrange the equation, grouping the terms with x together:
14x - 5 = 0.
2. Add the square of half the coefficient of x to both sides of the equation:
14x - 5 + (6/2)^2 = (6/2)^2.
14x - 5 + 9 = 9.
3. Simplify the equation:
14x + 4 = 9.
4. Add 5 to both sides of the equation:
14x + 4 + 5 = 9 + 5.
14x + 9 = 14.
5. Subtract 9 from both sides of the equation:
14x = 14 - 9.
14x = 5.
6. Divide both sides of the equation by 14 to solve for x:
x = 5/14.
So the root of the equation 8x + 6x - 5 = 0 is x = 5/14.