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Oksana_A [137]
3 years ago
14

Is 4 1/4 greater than or equal to 4 3/11

Mathematics
2 answers:
Oduvanchick [21]3 years ago
7 0
4 1/4 is greater than 4 3/11

Hope I helped!
~ Zoe
d1i1m1o1n [39]3 years ago
5 0
Actually it's less than
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Use the image to answer the question. Use the drop-down menus to complete the statements. It isfor Acacia to pick a purple piece
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Answer:

1. less likely

2. more likely

Step-by-step explanation:

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Solve the two-step equation. 14 = 31.7 – 3x
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14 = 31.7- 3x 

-17.7=-3x 

x=5.9
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An 11 lb turkey cost $15.00 while the 16 lb turkey is on sale for $20.00. How much would you save with the sale price? Please sh
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Find the base price per pound, $15/11, or $1.36 = 1lb, then multiply it by 16, 16 x 1.36 = $21.82. Now just subtract $20 from $21.82 = $1.82. So, You would save, on average, $1.82 by buying the sale.
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A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

4 0
3 years ago
Find the value of a in the parallelogram.
S_A_V [24]
6a-4 = 3a+54
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a = 58/3
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3 years ago
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