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3 years ago

What 2/3 + 5/7 complete denominators

2 answers:

8 0

You have to find common denominators. We can look at 3 and 7 and we know that nothing X 3 would equal to seven. So, we have to multiply
2/3 X 7 + 5/7 X 3
When we do that, we get 14/21 + 15/21
So when we add, the answer is
29/21
When we look at that, we have to see if it can be reduced. We know that both 7 and 3 can go into 21 but can they go into 29? No, so therefore, the answer cannot be reduced
Final answer is 29/21

dem82 [27]3 years ago

4 0

Multiplying numerators and denominators to get the LCD in all fraction denominators

2 × 7 + 5 × 3
------------------
3 × 7 + 7 × 3

Then rewriting the equation with the equivalent fractions

__14__ 15
= ----- + -----
__21 __21

With like denominators we can operate on just the numerators

__14 + 15
= ------------
___21

__29
= -------
__21

Simplifying the answer

__29
= -------
__21

___8
= 1 -------
___21

You might be interested in

O is the centroid. Find the value of x. (Sorry if u cant see the photo :/, but this is Geometry btw)

klemol [59]

Answer: x = 6

==============================================

Explanation:

The medians of a triangle meet up at the centroid in such a way that they cut each other at a ratio of 2:1, meaning that one part of the median is twice as long as the other. In this case, MO is two times longer than OP, so,

OP = 2*MO

and

MP = MO + OP

MP = MO + 2*MO .... replace OP with 2*MO

MP = 3*MO

9x-24 = 3*(x+4) ... plug in the given expressions

9x-24 = 3x+12

9x-3x = 12+24

6x = 36

x = 36/6

x = 6

8 0

3 years ago

Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d

Leni [432]

Answers:

(a) $y = \frac{1}{1 - Cx}$ , for any constant C

(b) Solution does not exist

(c) $y = \frac{256}{256 - 15x}$

(d) $y = \frac{64}{64 - 15x}$

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

$x\frac{dy}{dx} = y^2 - y
\\
\\ \indent xdy = \left ( y^2 - y \right )dx
\\
\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}
\\
\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} 
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}$ (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

$\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}
\\
\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} 
\\
\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }$ (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

$1 = (A+B)y - B
\\
\\ \indent \Rightarrow (A+B)y - B = 0y + 1
\\
\\ \indent \Rightarrow \begin{cases}
 A + B = 0
& \text{(3)}\\-B = 1
 & \text{(4)} \end{cases}$

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1

Hence,

$\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}$

So,

$\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} 
\\
\\ \indent \indent \indent \indent = \ln (y-1) - \ln y
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}$

Now, equation (1) becomes

$\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1
\\
\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2
\\
\\ \indent \frac{y-1}{y} = e^{C_1 - C_2}x
\\
\\ \indent \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}
\\
\\ \indent 1 - \frac{1}{y} = Cx
\\
\\ \indent \frac{1}{y} = 1 - Cx
\\
\\ \indent \boxed{y = \frac{1}{1 - Cx}}
$ (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

$

Hence, for any constant C, the following solution will pass thru (0, 1):

$\boxed{y = \frac{1}{1 - Cx}}$

(b) Using equation (5) in problem (a),

$y = \frac{1}{1 - Cx}$ (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that

$y = \frac{1}{1 - Cx}
\\
\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1$

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C.

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 16 = \frac{1}{1 - C(16)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 16C}
\\
\\ \indent 16(1 - 16C) = 1
\\ \indent 16 - 256C = 1
\\ \indent - 256C = -15
\\ \indent \boxed{C = \frac{15}{256}}


$

By replacing this value of C, the general solution becomes

$y = \frac{1}{1 - Cx}
\\
\\ \indent y = \frac{1}{1 - \frac{15}{256}x} 
\\ 
\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}
\\
\\
\\ \indent \boxed{y = \frac{256}{256 - 15x}}



$

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
- Substitute the values of x and y to the general solution.
- Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that

$y = \frac{1}{1 - Cx} 
\\ 
\\ \indent 16 = \frac{1}{1 - C(4)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 4C} 
\\ 
\\ \indent 16(1 - 4C) = 1 
\\ \indent 16 - 64C = 1 
\\ \indent - 64C = -15 
\\ \indent \boxed{C = \frac{15}{64}}$

Now, we replace C using the derived value in the general solution. Then,

$y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}$

5 0

3 years ago

How do you graph y= -5

olchik [2.2K]

Since y=k, there is a slope of zero, which is just a horizontal line k units from the x-axis. In this case k=-5, so the graph of y=-5 is a horizontal line 5 units below the x-axis...

7 0

3 years ago

SOMEONE PLEASE HELP ME!!!!! I NEED THIS!

snow_tiger [21]

Answer:

QR ≈ 16.3

Step-by-step explanation:

Since the figures are similar then the ratios of corresponding sides are equal, that is

$\frac{MN}{QR}$ = $\frac{MP}{QT}$ , substitute values

$\frac{40.3}{QR}$ = $\frac{57}{23}$ ( cross- multiply )

57 QR = 926.9 ( divide both sides by 57 )

QR ≈ 16.3 ( to the nearest tenth )

7 0

3 years ago

Which equation results from adding the equations in this system? 11 x minus 3 y = -17. Negative 4 x + 3 y = - 18.

marta [7]

Answer:

Therefore the results from adding the equation in this system is

$7x=-35$

Step-by-step explanation:

Given:

Equations as

$11x-3y=-17$ .......................1

$-4x+3y=-18$ .......................2

To Find :

Result when adding equation 1 and 2 = ?

Solution

On adding equation 1 and 2 the "3y" term will get cancel and the like terms will combine that is add (11x and -4x ) and ( -17 and - 18) as shown below

11x - 3y = -17

-4x + 3y = - 18

-------------------------------------------

7x + 0 = - 35

---------------------------------------------

Therefore the results from adding the equation in this system is

$7x=-35$

7 0

3 years ago

Read 2 more answers