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3 years ago

How do I solve for x? 2x + 16 = -5(x + 1)

1 answer:

4 0

P.E.M/D.A/S PEMDAS multiply subtract 2x from each side them its 16 = -7x -5 then add 5 to each side 21= -7x then divide x = -3

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Alana rode her bicycle 10 miles on Tuesday and 20 miles on Wednesday. What is the mean absolute deviation of this data?

stiks02 [169]

Answer:

Step-by-step explanation:

Given that :

Miles ridden in Tuesday = 10

Miles ridden on Wednesday = 20

Sample size, n = 2

Mean :

(10 + 20) / 2

= 30 /2

= 15

Mean absolute deviation :

(|10 - 15| + |20 - 15|) / 2

(5 + 5) / 2

10 / 2

= 5

3 0

2 years ago

Part C Now try this one. Write a description of the partitioned function using known function types, including transformations.

Sever21 [200]

Answer:

Following are the function description to the given question:

Step-by-step explanation:

In the given-question, three functions are used, that can be defined as follows:

In function 1:

This function is also known as the modulus on the absolute value function, for example:

$f(x)=| x| \left \{ {{x , \ \ \ x>0} \atop {-x, \ \ \ \x$

In the given in the above graph, that is $f(x) = -x , \ \ x$

In function 2:

In this function, It is an algebraic function that is $y=x^2$

It is also a part of the quadratic polynomial function, and its value is $y=x^2 , \ \ \ x> 0$

In function 3:

In this function, it is the cubic polynomial equation that's value is $y=x^3$

In the graph its value is:

$y=-x^3\\\\and \\ \\\to y= f(x) \\\\ \to y=-f(x)\\$

6 0

2 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago

]Choose the equation below that represents the line passing through the point (-2, -3) with a slope of -6. . . y + 3 = -6(x + 2)

pshichka [43]

Y - y1 = m(x - x1)
slope(m) = -6
(-2,-3)...x1 = -2 and y1 = -3
now sub and pay close attention to ur signs
y - (-3) = -6(x - (-2)....we're not done yet...
y + 3 = -6(x + 2) <===

7 0

3 years ago

the average price of a gallon of gas was $3.22 and 2014 and $2.40 in 2015 what is the percent decrease in the price of gas

kakasveta [241]

Answer:

25.47%

Step-by-step explanation:

6 0

3 years ago