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attashe74 [19]
2 years ago
12

The current temperature in Smalltown is 20°F. This is 6 degrease less than twice the temperature that it was six hours ago. What

was the temperature in Smalltown six hours ago?
Mathematics
1 answer:
jeyben [28]2 years ago
7 0

Answer:

<em>The temperature six hours ago was 13 °F</em>

Step-by-step explanation:

Six hours ago the temperature was x degrees.

The temperature now is "6 degrees less than twice the temperature that it was six hours ago."

The temperature now is 2x - 6.

The temperature now is 20 deg.

2x - 6 = 20

2x = 26

x = 13

Answer: The temperature six hours ago was 13 °F.

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2 years ago
Suppose that a rectangular picture is surrounded by a frame of uniform width of x
user100 [1]

Answer:

Ok so the area of the frame is 18*12 = 216cm^2

Let’s suppose the width of the frame is x

The total area (painting and frame) is (2x+18)(2x+12) and this is equal to 432

4x^2 + 60x + 216 = 432

4x^2 + 60x - 216 = 0

x^2 + 15x - 54 = 0

(x+18)(x-3) = 0

Therefore x = {-18,3}

Since width of the frame has to be positive, the width has to be 3cm Step-

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2 years ago
Divide: 4/0 is it 0, undefined, 1, 4 or none of the above?
Gennadij [26K]

Answer: Undefined

Step-by-step explanation:

4/0 is undefined. This is because any number divided by zero is undifined.

7 0
3 years ago
Read 2 more answers
Is that right? <br><br> If not please help me out.
Alex73 [517]

Answer:

Yeah.it is right

Step-by-step explanation:

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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
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