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3 years ago

Which expression represents the number of feet that are in x yards?

Mathematics

2 answers:

nignag [31]3 years ago

7 0

C

I think that this is the correct answer

AleksandrR [38]3 years ago

3 0

we know that

$1\ yard=3\ feet$

Let

y-------> the number of feet that are in x yards

by proportion

<u>Find the value of y</u>

$\frac{3}{1} \frac{feet}{yard}=\frac{y}{x} \frac{feet}{yards} \\ \\y=3x\ feet$

therefore

<u>the answer is the option C</u>

$3x$

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A line that includes the point (-9, 1) has a slope of 1. What is its equation in

SVETLANKA909090 [29]

Answer:

y=x+10

Step-by-step explanation:

1.) y-1=1(x+9)

2.) y-1=x+9

3.) y=x+10

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3 years ago

What is the slope-intercept form in the picture

grandymaker [24]

Answer:

y= 3/2x+40

Step-by-step explanation:

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3 years ago

If DF=78, DE=5x-9, and EF=2x+10, find DE.

Kazeer [188]

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

<h3>What is the numerical value of DE?</h3>

Given the data in the question;

E is a point between point D and F.
Segment DF = 78
Segment DE = 5x - 9
Segment EF = 2x + 10
Numerical value of DE = ?

Since E is a point between point D and F.

Segment DF = Segment DE + Segment EF

78 = 5x - 9 + 2x + 10

78 = 7x + 1

7x = 78 - 1

7x = 77

x = 77/7

x = 11

Hence,

Segment DE = 5x - 9

Segment DE = 5(11) - 9

Segment DE = 55 - 9

Segment DE = 46

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

Learn more about equations here: brainly.com/question/14686792

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3 0

1 year 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

Misha Larkins [42]

Answer:

It's 130.66, D

Step-by-step explanation:

:)) yw

8 0

3 years ago

Read 2 more answers