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3 years ago

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Sam has $140 to shop with… He wants to buy a pair of jeans that cost $45 and then some rings. The Rings cost $16 each. If Sam pu

rchases the Jeans first, what is the maximum number of rings that Sam can purchase?

1 answer:

m_a_m_a [10]3 years ago

6 0

Answer:

5 rings

Step-by-step explanation:

sam can only buy 5 rings because each ring cost $16 and after he bought the jeans which cost $45 it left him with only $95. so if you times 16 and 5 it gives you 80 and if you times 16 and 6 it gives you 96 but sam only had $95 left so he could only buy 5 which cost him $80. that left him with an extra $15.

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DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

<u><em>the answer is </em></u>

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

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$x^{2} -4x-7x+28$

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$x(x-4)-7(x-4)$

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and -12x can be rewritten as $-2*3x*2$

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There's a rule that says $a^{2} -2ab+b^{2} = (a-b)^{2}$

In our case, a=3x and b=2

<u><em>so the final answer is</em></u>

$(3x-2)^2$

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Step-by-step explanation:

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