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chubhunter [2.5K]
3 years ago
9

Write 25 and 13 as equivalent fractions using a common denominator.

Mathematics
2 answers:
Harman [31]3 years ago
8 0

Answer:

25/50 + 13/26

Step-by-step explanation:?????Not sure, please have someone else answer i dont want to be the cause of your incorrect answer:(

beks73 [17]3 years ago
5 0
<h2><u><em>important: 25 13 </em></u></h2><h2><u><em> looks like a fraction, but it is actually an improper fraction. </em></u></h2><h2 /><h2><u><em>There is an infinity number of equivalent fractions to 25 13 </em></u></h2><h2><u><em>To find an equivalent fraction to 25  13  or to any other fraction, you just need to multiply (or divide, if the fraction is not yet reduced), both the numerator and the denominator of the given fraction by any non-zero natural number. </em></u></h2><h2><u><em></em></u></h2><h2><u><em>For example: By multiplying the original fraction by 2, we get: </em></u></h2><h2><u><em>25 × 2 </em></u></h2><h2><u><em>13 × 2 </em></u></h2><h2><u><em> = 50 </em></u></h2><h2><u><em>26</em></u></h2><h2 /><h2><u><em>Here is the full list of equivalent fractions to 25 \13 </em></u></h2><h2><u><em>25/13, 50/26, 75/39, 100/52, 125/65, 150/78, 175/91, 200/104, 225/117, 250/130, 275/143, 300/156, 325/169, 350/182, 375/195, 400/208 425/221, 450/234, 475/247, 500/260 ...</em></u></h2>

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A scooter can travel between 22 and 28 miles on each gallon of gasoline.If one gallon of gasoline costs between $3.75 and 3.95 p
just olya [345]

Answer:

Cost to travel 75 miles = 11.55 $

Explanation:

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 One gallon of gasoline costs between $3.75 and 3.95 per gallon.

Average cost for one gallon of gasoline = (3.75+3.95)/2 = 3.85 $

We need to find cost for traveling 75 miles.

Gasoline required = Total miles/ Average distance a scooter can travel with each gallon of gasoline

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Cost to travel 75 miles = Gasoline required x Average cost for one gallon of gasoline

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3 years ago
What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
a_{n+1}=a_{n-3}+4d

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

a_{n+1}=a_1+nd

or, shifting the index by one so that the formula gives the nth term explicitly,

a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

a_{n+1}=ra_n

Well, since a_n is just the term after a_{n-1} scaled by r, you can write

a_{n+1}=r(ra_{n-1})=r^2a_{n-1}

Doing this again and again, you'll see a similar pattern emerge:

a_{n+1}=r^2a_{n-1}
a_{n+1}=r^2(ra_{n-2})
a_{n+1}=r^3a_{n-2}
a_{n+1}=r^3(ra_{n-3})
a_{n+1}=r^4a_{n-3}

and so on. Notice that the subscript and the exponent of the common ratio both add up to n+1. For instance, in the third equation, 3+(n-2)=n+1. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

a_n=r^{n-1}a_1
6 0
4 years ago
Joseph is standing 12 feet from a mirror lying on the ground, and his eyes are 5 feet above the ground. The line-of-sight reflec
eimsori [14]
See the attached figure which represents the explanation of the problem.
Let <span>the height of the building = h
</span>
∵ ∠1 congruent to ∠2
∴ ∠1 = ∠2
∴ tan (∠1) = tan (∠2)    ⇒⇒⇒  tan of the angle =  (opposite/adjacent)

∴ \frac{5}{12} =  \frac{h}{264}
∴ 12 h = 5 * 264 = 1,320
∴ h = 1,320/12 = 110 feet

∴ T<span>he height of the building = 110 feet
</span>

8 0
3 years ago
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