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Digiron [165]
3 years ago
7

A charity has received donations totaling $2370. However, this is only three-fifths of its goal.

Mathematics
2 answers:
Alinara [238K]3 years ago
8 0
3,950 is your answer. See, you divide 2,370 by 3 to get 1/5 of their goal and then multiply that by 5.
topjm [15]3 years ago
5 0
\sf2350=\frac{3}{5}x\\\\Multiply\ by\ 5\ on\ both\ sides.\\2350\times5=3x\\11750=3x\\\\Divide\ by\ 3\ on\ both\ sides.\\\\{\boxed{x=3950\ dollars}
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Solve the equation 5y-6=10
BaLLatris [955]

Answer:

y = 3.2

Step-by-step explanation:

Simplifying

5y + -6 = 10

Reorder the terms:

-6 + 5y = 10

Solving

-6 + 5y = 10

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Add '6' to each side of the equation.

-6 + 6 + 5y = 10 + 6

Combine like terms: -6 + 6 = 0

0 + 5y = 10 + 6

5y = 10 + 6

Combine like terms: 10 + 6 = 16

5y = 16

Divide each side by '5'.

y = 3.2

Simplifying

y = 3.2

8 0
2 years ago
Read 2 more answers
Jack earns $50 for washing cars on the weekdays He makes an average of $4.25 in tips per hour right the function of Jacks earnin
kobusy [5.1K]

Answer:

Function: y= $4.25x + $50; After 6 hours: $25.50

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
SOMEONE HELP ME PLEASE
IgorC [24]

A standard die has numbers = 1,2,3,4,5 and 6

Total outcomes = 6

P = Number of outcomes/Total Outcomes

Even and less than 4 = 1,2,3,4 and 6

P = 5/6

Answered by GauthMath if you like please click thanks and comment thanks

3 0
2 years ago
A King wanted to replace his Prime-Minister but didn't want to upset him too much. So he called the Prime-Minister to his chambe
Neporo4naja [7]

Answer: Ask the king to draw first and read it. Explain that if the king selects "leave" the PM's choice could only be "stay". It is then unnecessary for the PM to draw. It avoids embarrassing the king in his lie, demonstrates the PM's intelligence, and keeps his job.

Step-by-step explanation:

3 0
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
3 years ago
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