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3 years ago

Pierce is making a rectangular frame for a photo collage that has perimeter of

Mathematics

1 answer:

Ymorist [56]3 years ago

3 0

Answer:

sure

Step-by-step explanation:

please mark me as the brainliest

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Mrs. Johnson put 22,250 miles

Viefleur [7K]

The change in quantities can be represented as percentage

<em>The percentage decrease in miles driven is 34.83%</em>

The given parameters are:

$\mathbf{Initial = 22250}$

$\mathbf{Final = 14500}$

The percentage decrease is calculated as:

$\mathbf{\%Decrease = \frac{Final - Initial}{Initial} \times 100\%}$

Substitute known values

$\mathbf{\%Decrease = \frac{14500- 22250}{22250} \times 100\%}$

$\mathbf{\%Decrease = \frac{-7750}{22250} \times 100\%}$

$\mathbf{\%Decrease = \frac{-775000}{22250} \%}$

$\mathbf{\%Decrease = -34.83 \%}$

The negative sign implies that, the change represents a decrement.

Hence, the percentage decrease in miles driven is 34.83%

Read more about percentage change at:

brainly.com/question/14801224

7 0

2 years ago

List the values that make up the domain of this relation.

balandron [24]

$\bf (\stackrel{\downarrow }{5},3),(\stackrel{\downarrow }{2},-7),(\stackrel{\downarrow }{1},-8),(\stackrel{\downarrow }{8},3)\qquad \quad \stackrel{\textit{the domain is always the set of the 1st coordinate}}{(5,2,1,8)}$

6 0

3 years ago

(b) factorise <br>(3m-1)(6-a)-(m+3)(6-a)

Step2247 [10]

Answer:

Step-by-step explanation:

(3m-1)(6-a)-(m+3)(6-a)

=3m(6-a)-1(6-a)-m(6-a)+3(6-a)

=18m-3ma-6+a-6m+ma+18-3a

=12m-2ma+12-2a

=2(6m-ma+6-a)

7 0

3 years ago

NEED HELP ASAP!!

Kaylis [27]

Answer:

20²+21² = x²

841= x²

x= 29

nearest tenth = 30

4 0

3 years ago

Read 2 more answers

A software designer is mapping the streets for a new racing game. all of the streets are depicted as either perpendicular or par

ElenaW [278]

The equation of the central street PQ is -1.5x - 3.5y = -31.5 option (b) is correct.

<h3>What is a straight line?</h3>

A straight line is a combination of endless points joined on both sides of the point.

We have a straight line:

$-7x+3y=-21.5$

Convert it to the general form given below:

$\rm y=mx+c$

$\rm 3y=-21.5+7x\\\\ \rm y = \frac{-21.5}{3}+\frac{7}{3}x$ or

$\rm y = \frac{7}{3}x-\frac{21.5}{3}$

$m = \frac{7}{3}$ (Slope of AB line)

For the slope(m') of the PQ line:

$\rm m'=-\frac{1}{m}$ ( because AB and PQ are perpendicular to each other)

$\rm m' = -\frac{3}{7}$

We know the:

$\rm (y-y')=m'(x-x')$

Where (x', y') = (7, 6), we get:

$\rm (y-6)=-\frac{3}{7} (x-7)$

$\rm 7(y-6)=-3 (x-7)\\\\\rm 7y-42= -3x+21\\\\\rm 7y= -3x+21+42\\\\\rm 3x+7y=63$

$\rm -1.5x-3.5y=-31.5$ (multiply by -1/2 on both sides)

Thus, the equation of the central street PQ is -1.5x - 3.5y = -31.5

Learn more about the straight line.

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6 0

2 years ago