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3 years ago

BRAINLIESTTT ASAP!!! PLEASE ANSWER :) Explain your answer.

Mathematics

2 answers:

baherus [9]3 years ago

6 0

Answer:

x=14

Step-by-step explanation:

2(3x-4)=5x+6

first get rid of the parentheses by multiplying the 2

6x-8 = 5x+6

put like terms on one side of the equation. Start with the 5x, move it to the left side

(6x-8)-5x = (5x+6)-5x

x-8 = 6

Again, put like terms on one side of the equation. Move the 8 to the right side

(x-8)+8 = 6+8

=> x = 14

4x - 3 = 2(x-1)

Start again by getting rid of the parentheses to make it a clean equation... aka distribute the 2:

4x-3 = 2x - 2

Move the like terms to one side of the equation. Start with x.. move 2x to the left:

(4x-3)-2x = (2x-2)-2x

=> 2x-3 = -2

Move the like terms to one side of the equation.Now move the -3 to the other side:

(2x-3)+3 = -2+3

=> 2x=1

Lastly, divide by 2 on both sides to find out what x is:

(2x)/2=1/2

=> x = 1/2

..so no, Charlie was incorrect.

Evgesh-ka [11]3 years ago

5 0

Answer:

x=14

In the second line, he changed the sign on the 2

Step-by-step explanation:

2(3x-4) = 5x+6

Distribute the 2

6x-8 = 5x+6

Subtract 5x from each side

6x-5x -8 = 5x-5x+6

Add 8 to each side

x-8+8 = 6+8

x=14

b. 4x-3 = 2(x-1)

Distribute

4x-3 =2x-2 He changed the sign on the 2

Subtract 2x from each side

4x-2x -3 = 2x-2x-2

2x-3 =-2

Add 3 to each side

2x-3+3 =-2+3

2x =1

Divide by 2

2x/2 =1/2

x=1/2

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Plz help me this was due yestrday but my mom wont let me do it

AURORKA [14]

Answer:

120

Step-by-step explanation:

z=8 w=8

7(8)=56

8(8)=64

56+64=120

8 0

3 years ago

Find the missing lengths of the sides

dusya [7]

Answer:

a=9, c=19

Step-by-step explanation:

Aight so basically in 30, 60, 90 triangles there is a formula for the sides

the hypotenuse(the longest side)=2a

The tiny side (the smallest side)=a

and the middle child = $a\sqrt{3}$

Based on that we see that a is 9

and since c=2a then it's 18

4 0

3 years ago

Read 2 more answers

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

Help me again please.

PtichkaEL [24]

Answer:

$m=\frac{8}{5}$

Step-by-step explanation:

7 0

3 years ago

A fish swam 75 feet to the bottom of the river. The fish then swam up 35 feet towards the surface of the river and stopped. How

telo118 [61]

Answer: 40

Step-by-step explanation:

75-35

5 0

3 years ago