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3 years ago

Find the equivalent expression for x2 + 4x + 4

Mathematics

2 answers:

Vlada [557]3 years ago

4 0

Step-by-step explanation:

Add.

2x+4x=6x

6x+4=6x+4

dlinn [17]3 years ago

4 0

Answer:

6x+4=6x+4 Step-by-step explanation:

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Find the circumstances of the circle

Mkey [24]

The meaning of circumference is to find the distance around a circle or in other words perimeter by using the formula C=2πr or C=πd.

On this case it is asked you to find the circumference of a circle with a diameter of 3 meters, so you will use the formula C=2πr.

C=2πr
C=2(3.14)(3)
C=18.84 meters

The circumference of the given circle is 18.84 meters.

8 0

3 years ago

Can someone please help me with the first 2 questions?

Ber [7]

Answer:

a) 10.47

b) 10.23

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2 years ago

Read 2 more answers

Suppose that a cyclist began a 476

mojhsa [17]

Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards

the distance between them is 476 miles, and they met 8.5hrs later

let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs

we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2

thus $\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{eastbound cyclist}&d&r&8.5\\
\textit{westbound car}&476-d&r+33.2&8.5
\end{array}\\\\
-----------------------------\\\\
\begin{cases}
\boxed{d}=8.5r\\\\
476-d=(r+33.2)8.5\\
----------\\
476-\boxed{8.5r}=(r+33.2)8.5
\end{cases}$

solve for "r", to see how fast the cyclist was going

what about the car? well, the car's rate is r + 33.2

6 0

3 years ago

What is the value of the expression below when z = 4? 10z - 3

sergey [27]

Answer:

Your correct answer is 37

Step-by-step explanation:

Evaluate for z = 4

(10)(4)−3

= 37

7 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