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Ksju [112]
3 years ago
11

Four gallons of gasoline weigh 25 pounds . find the unit rate in pounds per gallon.

Mathematics
2 answers:
Papessa [141]3 years ago
8 0

Answer:

6.25 pounds

Step-by-step explanation:

\frac{25}{4} = \frac{x}{1}

25÷4 = 6.25

\frac{25}{4} = 6.25

notka56 [123]3 years ago
7 0
There are several information's that are already given in the question. Based on those information's the answer to the question can be easily deduced. 
Weight of 4 gallons of gasoline = 25 pounds
Then
Weight of 1 gallon of gasoline = 25/4 pounds
                                                = 6.25 pounds
From the above deduction, we can easily conclude that the unit rate in pounds per gallon of gasoline is 6.25 pounds/gallon. I hope the answer helps you.
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Answer:

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Step-by-step explanation:

if you need them added up here: 4+8 =12 12+16 = 28

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5a + 12d - 19a -11d -a
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-15a+d

Step-by-step explanation:

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Evaluate the integral. (Assume a ≠ b. Remember to use absolute values where appropriate. Use C for the constant of integration.)
Bess [88]

Answer:

<u><em>F(x)= 5*[\frac{x^{3} }{3} + (a*b)*\frac{x^{2} }{2} + a*b*x + C.</em></u>

Step-by-step explanation:

<u><em>First step we aplicate distributive property to the function.</em></u>

<u><em>5*(x+a)*(x+b)= 5*[x^{2}+x*b+a*x+a*b]</em></u>

<u><em>5*[x^{2}+x*(b+a)+a*b]= f(x), where a, b are constant and a≠b</em></u>

<u><em>integrating we find ⇒∫f(x)*dx= F(x) + C, where C= integration´s constant</em></u>

<u><em>∫^5*[x^{2}+x*(a+b)+a*b]*dx, apply integral´s property</em></u>

<u><em>5*[∫x^{2}dx+∫(a*b)*x*dx + ∫a*b*dx], resolving the integrals </em></u>

<u><em>5*[\frac{x^{3} }{3} + (a*b)*\frac{x^{2} }{2} + a*b*x</em></u>

<u><em>Finally we can write the function F(x)</em></u>

<u><em>F(x)= 5*[\frac{x^{3} }{3} + (a*b)*\frac{x^{2} }{2} + a*b*x ]+ C.</em></u>

4 0
3 years ago
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