If the pattern below was tripled, what would be the ratio of black squares to gray squares

1 answer:

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If the pattern (see attachment) below was tripled, the ratio of black squares to gray squares would be equal to 1:2.

<h3>What is a proportion?</h3>

A proportion can be defined as an equation which is typically used to represent (indicate) the equality of two (2) ratios. This ultimately implies that, proportions can be used to establish that two (2) ratios are equivalent and solve for all unknown quantities.

Mathematically, a direct proportion can be represented by the following equation:

y = kx

Where:

y and x are the variables.
k represents the constant of proportionality.

<h3>What is a ratio?</h3>

A ratio can be defined as a mathematical expression that's used to denote the proportion of two (2) or more quantities with respect to one another and the total quantities.

For this exercise, the ratio of black square to gray squares is given by:

B:G = 1:2

Tripling the pattern, we would multiply by three (3) as follows:

B:G = [1:2] × 3

B:G = 3:6

Dividing all through by 3, we have:

B:G = [3:6]/3

B:G = 3/3 : 6/3

B:G = 1:2.

Read more on ratio here: brainly.com/question/13513438

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When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance

bekas [8.4K]

a) $v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

The average speed is equal to the ratio between the total distance ( $d_{tot}$ and the total time taken ( $t_{tot}$ ):

$v=\frac{d_{tot}}{t_{tot}}$

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

$d_1 = (3 h)(60 mph)=180 mi$

So the total distance is

$d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi$

The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

$t_{tot}=3 h +t_2$

So, the equation can be rewritten as:

$v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

b) 0.50 h (half a hour)

Since we know the value of the average speed, $v=57.14 mph$ , we can substitute it into the previous equation to find the value of $t_2$ , the time the trucker drove in the city: