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3 years ago

Margo can purchase tile at a store for $0.79 per tile and rent a tile saw for $10. At another store she can

Mathematics

1 answer:

PolarNik [594]3 years ago

8 0

If she buys 60 tiles, the cost at both shops is the same.

If she buys less than 60 tiles, then the second shop is cheaper.

If she buys more than 60 tiles, then the first shop is cheaper.

Explanation:

Let the number of tiles be

At the first shop: Cost =

0.79

At the second shop: Cost =

1.19

If the cost is the same:

1.19

0.79

←

solve for x

1.19

−

0.79

0.4

tiles

If she buys less than 60 tiles, then the second shop is cheaper.

If she buys more than 60 tiles, then the first shop is cheaper.

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Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

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r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $100

r = 11% = 11/100 = 0.11

n = 1 because it was compounded once in a year.

t = 6 years

Therefore,.

A = 100(1 + 0.11/1)^1 × 6

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Step-by-step explanation:

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3 years ago

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The table shows values for functions f(x) and g(x). x f(x)=2x−3 g(x)=73x−2 −1 −52 −133 0 −2 −2 1 −1 13 2 1 83 3 5 5 4 13 223 5 2

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f(x) = g(x) will be the input or x value at which f and g have the same output or y value. Look in the table where two numbers repeat right next to each other.

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3 years ago

Kate bought four Gatorade for $2 each, as well as twenty four Cup of Noodles. If her total was $26, how

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4 years ago

If the numbers 4, 5 and 6 are each used exactly once to replace the letters in the expression A ( B − C ), what is the least pos

guapka [62]

Answer:

The least possible result is <em>-10</em>.

Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

First of all,

Let A = 4, B = 5 and C = 6

$A(B-C) = 4 \times (5-6) = 4 \times -1 = -4$

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$A(B-C) = 4 \times (6-5) = 4 \times 1 = 4$

Let A = 5, B =4 and C = 6

$A(B-C) = 5 \times (4-6) = 5 \times -2 = -10$

Let A = 5, B = 6 and C = 4

$A(B-C) = 4 \times (6-4) = 4 \times 2 = 8$

Let A = 6, B = 4 and C = 5

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$A(B-C) = 6 \times (6-5) = 6 \times 1 = 6$

Summarizing the above values in the form of a table:

$\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\ 4 & 6 & 5 & 4\\ 5 & 4 & 6 & -10\\ 5 & 6 & 4 & 10\\ 6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}$

So, the least possible result is <em>-10</em>.

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