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Colt1911 [192]
3 years ago
8

How do u simplify 5/4 as fraction

Mathematics
2 answers:
mafiozo [28]3 years ago
8 0
Make it a mixed number 1 1/4
charle [14.2K]3 years ago
8 0
It would be 1 and 1/4
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alexandr402 [8]

Answer:

The answer is E

Step-by-step explanation:

7 0
3 years ago
7.A sound system has a regular price of $249. Find the total cost if it is on sale for 50% off, and the sales tax is 5.75$.
Nady [450]

Answer:

<u>$131.66</u>

Step-by-step explanation:

50 percent off 249.00 is 124.50

Then 124.50 including a sales tax of 5.75$

Would Be <u>$131.66</u>

------------------------------------------------------------------

<em><u>Hope this Helps!!! :)</u></em>

If you have any more questions feel free to ask!

5 0
2 years ago
In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with
Zarrin [17]

First check the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

y_c = C_1e^{-2t} + C_2te^{-2t}

For the particular solution corresponding to 12te^{-2t}, we might first try the <em>ansatz</em>

y_p = (At+B)e^{-2t}

but e^{-2t} and te^{-2t} are already accounted for in the characteristic solution. So we instead use

y_p = (At^3+Bt^2)e^{-2t}

which has derivatives

{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}

{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}

Substituting these into the left side of the ODE gives

(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.

For the second solution corresponding to -8t-12, we use

y_p = Ct + D

with derivative

{y_p}' = C

{y_p}'' = 0

Substituting these gives

4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12

so that 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -1.

Then the general solution to the ODE is

y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have

-2 = C_1 - 1 \implies C_1 = -1

1 = -2C_1 + C_2 - 2 \implies C_2 = 1

and so the particular solution satisfying these conditions is

y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1

or

\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}

7 0
2 years ago
What is the equations for f(x)?
Wittaler [7]

f(x) = -8\cdot 4^x


You reflect a function with respect to the x axis by changing its sign.


In fact, if you consider the two alternatives


x \mapsto f(x) and x \mapsto -f(x)


you can see that you're choosing opposite outputs for the same inputs.


In other words, one graph will be made of points like (x,f(x)), the other of points like (x,-f(x))


And the transformation


(x,y) \to (x,-y)


is exactly a reflection with respect to the x axis.

6 0
3 years ago
X + x(x + y) + y; use x = 4, and y = 1
leva [86]

\huge\text{Hey there!}

\huge\text{x + x(x + y) + y}

\huge\text{= 4 + 4(4 + 1) + 1}

\huge\text{= 4 + 4(5) + 1}

\huge\text{= 4 + 20 + 1}

\huge\text{= 24 + 1}

\huge\text{= 25}

\huge\boxed{\textsf{Therefore, your answer is: 25}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}

~\huge\boxed{\frak{Amphitrite1040:)}}

5 0
2 years ago
Read 2 more answers
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