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1 year ago

If she chooses the 8-inch tiles, she will need a quarter as many tiles as she would with 2-inch tiles.

1 answer:

8 0

In the above word problem, If she chooses the 8-inch tiles, she will need a quarter as many tiles as she would with 2-inch tiles, Quarter 8-inch tiles will cover the same area as one 2-inches.

<h3>What is the justification for the above?</h3>

Note that the area of the one 2-inch tiles is given as:

A1 = 4in²

The area of the quarter 8-inch tiles is:

A2 = 1/4 x 8 x 8

A2 = 16inch²

Divide both areas

A2/A1= 16/4

= 4

This implies she'll need four 2-inch tiles to cover the same amount of space as a quarter 8-inch tile.

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Full Question:

A homeowner is deciding on the size of tiles to use to fully tile a rectangular wall in her bathroom that is 80 inches by 40 inches. The tiles are squares and come in three side lengths: 8 inches, 4 inches, and 2 inches. State if you agree with each statement about the tiles. Explain your reasoning.

If she chooses the 8-inch tiles, she will need a quarter as many tiles as she would with 2-inch tiles,

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Read 2 more answers

What is the exact time t, in years, needed for the balance of an account that earns 5% annual 8. Interest compounded continuousl

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It takes 22.52 years for the balance to triple in value.

Step-by-step explanation:

Continuous compounding:

The amount of money earned using continuous compounding is given by the following equation:

$A(t) = A(0)(1+r)^t$

In which A(0) is the initial amount of money and r is the interest rate, as a decimal.

Interest rate of 5%.

This means that $r = 0.05$ , and thus:

$A(t) = A(0)(1+r)^t$

$A(t) = A(0)(1+0.05)^t$

$A(t) = A(0)(1.05)^t$

Time for the balance to triple?

This is t for which $A(t) = 3A(0)$ . So

$A(t) = A(0)(1.05)^t$

$3A(0) = A(0)(1.05)^t$

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$t = \frac{\log{3}}{\log{1.05}}$

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3 years ago

I don’t understand this question?

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3 years ago

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

charle [14.2K]

ANSWER

See below

EXPLANATION

Given

$f(x) = \frac{ {x}- 9 }{x + 5}$

and

$g(x) = \frac{ - 5x - 9}{x - 1}$

$(f \circ \: g)(x)= \frac{ (\frac{ - 5x - 9}{x - 1})- 9 }{(\frac{ - 5x - 9}{x - 1} )+ 5}$

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9 - 9(x - 1)}{x - 1}}{\frac{ - 5x - 9 + 5(x - 1)}{x - 1} }$

Expand:

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9 - 9x + 9}{x - 1}}{\frac{ - 5x - 9 + 5x - 5}{x - 1} }$

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9x + 9 - 9}{x - 1}}{\frac{ - 5x + 5x - 5 - 9}{x - 1} }$

$(f \circ \: g)(x)= \frac{ \frac{ - 14x }{x - 1}}{\frac{ -14}{x - 1} }$

Since the denominators are the same, they will cancel out,

$(f \circ \: g)(x)= \frac{ - 14x}{ - 14} = x$

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3 years ago