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Lunna [17]
4 years ago
7

Answer this correctly

Mathematics
1 answer:
Elis [28]4 years ago
4 0

Answer:

1. 0.56=14/25

2. 0.45=9/20

3. 6.02=301/50

4. 1.4=7/5

5. 0.12= 3/25

Step-by-step explanation

REALLY EASY... IF U NEED MORE HELP LET ME KNO

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Diana is working on a painting for school. She needs 1 1/5 ounces of yellow paint and 2 1/2 ounces of orange paint. How much les
Eduardwww [97]

Answer:

1 3/10

Step-by-step explanation:

We are to determine the difference the paint of yellow and orange paint

The equation we are to solve is

2\frac{1}{2}  - 1\frac{1}{5}

The equation can be solved by taking the following steps

  1. Subtract the two whole numbers
  2. divide each denominator by the least common multiple of the denominator and multiply the figures by the numerator

1\frac{5 - 2}{10} = 1\frac{3}{10}

3 0
3 years ago
Hayley picked 2/3 of a pound of
Ksenya-84 [330]
The answer the this is 0.23 pounds per container
8 0
3 years ago
Prove that the two circle shown below are similar 9.01
liraira [26]

Answer:

If you draw the linear line to their similar points and set equal then solve you can see their similar.... also can someone come help me answer my question? I NEED HELP PLS ANYONE!!

Step-by-step explanation:

4 0
4 years ago
Write the equation of the line with a slope of -1/3 and passing through the point (6, -4).
Lina20 [59]
Y = -1/3x + b
then plug it in
-4 = (-1/3 x 6) + b. (6 times -1 divided by 3)
-4 = -2 + b. (move -2 to other side)
b = -2. (now put em all together)
y = -1/3x -2
4 0
3 years ago
Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.
Jlenok [28]
y''+2y'+y=2t^2-t+3e^{-2t}

The characteristic equation is

r^2+2r+1=(r+1)^2=0\implies r=-1

so the characteristic solution is

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

For the particular solution, we can try looking for a solution of the form

y_p=at^2+bt+c+de^{-2t}
\implies{y_p}'=2at+b-2de^{-2t}
\implies{y_p}''=2a+4de^{-2t}

Substituting into the ODE, we have

(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}
at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}
\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3

So the general solution to the ODE is

y=y_c+y_p
y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}
4 0
3 years ago
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