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Andrew [12]
3 years ago
5

Express each ratio as a unit rate. Round to the nearest tenth or cent. 

Mathematics
1 answer:
Mariana [72]3 years ago
8 0

Answer:

1=0.84 per gallon

2=5.25 per lb

Step-by-step explanation:

Divide the cost or amount by how many of what there is. Ex= 10.5/2=5.25

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What is the simplified value of the Exponential expression . 16 to the power of 1/4
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Step-by-step explanation:

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4 years ago
The first term of a geometric sequence is 6 and the fourth term is 384. What is the sum of the first 10 terms of the correspondi
aleksandrvk [35]

Answer:

2097150

Step-by-step explanation:

<u>GIVEN :-</u>

  • First term of G.P. = 6
  • Forth term of G.P. = 384

<u>TO FIND :-</u>

  • Sum of first 10 terms of the G.P.

<u>CONCEPT TO BE USED IN THIS QUESTION :-</u>

<em>Geometric Progression :-</em>

  • It's a sequence in which the successive terms have same ratio.
  • General form of a G.P. ⇒ a , ar , ar² , ar³ , ....... [where a = first term ; r = common ratio between successive terms]
  • Sum of 'n' terms of a G.P. ⇒ \frac{a(r^n - 1)}{r-1}.

<em>[NOTE :- </em>\frac{a(1-r^n)}{1-r} can also be the<em> formula for "Sum of n terms of G.P." because if you put 'r' there (assuming r > 0) you'll get negative value in both the numerator &  denominator from which the negative sign will get cancelled from the numerator & denominator. </em><em>YOU'LL BE GETTING THE SAME VALUE FROM BOTH THE FORMULAES.</em><em>]</em>

<u>SOLUTION :-</u>

Let the first term of the G.P. given in the question be 'a' and the common ratio between successive terms be 'r'.

⇒ a = 6

It's given that <u>forth term</u> is 384. So from "General form of G.P." , it can be stated that :-

=> ar^3 = 384

=> 6r^3 = 384

Divide both the sides by 6.

=> \frac{6r^3}{6} = \frac{384}{6}

=> r^3 = 64

=> r = \sqrt[3]{64} = 4

Sum of first 10 terms = \frac{6(4^{10}-1)}{4 - 1}

                                     = \frac{6(1048576 - 1)}{3}

                                     = 2 \times 1048575

                                     = 2097150

6 0
3 years ago
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