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1 year ago

Which equation represents a line that passes through (-2, 4) and has a slope of 2/5

Mathematics

1 answer:

Ronch [10]1 year ago

3 0

An equation represents a line that passes through (-2, 4) and has a slope of 2/5 is y-4 = 2/5(x+2)

<h3>Equation of a line</h3>

The equation of a line in point-slope form is expressed as;

y-y1 = m(x - x1)

where

m is the slope

(x1, y1) is any pint on the line

Substitute the given parameter

y - 4 = 2/5(x-(-2))

y-4 = 2/5(x+2)

Hence an equation represents a line that passes through (-2, 4) and has a slope of 2/5 is y-4 = 2/5(x+2)

Learn more on equation of a line here: brainly.com/question/13763238

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2 years ago

4C. Quintin is using the three different shaped

Sauron [17]

The smallest number of tiles Quintin will need in order to tile his floor is 20

The given parameters;

number of different shapes of tiles available = 3

number of each shape = 5

area of each square shape tiles, A = 2000 cm²

length of the floor, L = 10 m = 1000 cm

width of the floor, W = 6 m = 600 cm

To find:

the smallest number of tiles Quintin will need in order to tile his floor

Among the three different shapes available, total area of one is calculated as;

$A_{one \ square \ type} = 5 \times 2000 \ cm^2 = 10,000 \ cm^2$

Area of the floor is calculated as;

$A_{floor} = 1000 \ cm \times 600 \ cm = 600,000 \ cm^2$

The maximum number tiles needed (this will be possible if only one shape type is used)

$maximum \ number= \frac{Area \ of \ floor}{total \ area \ of \ one \ shape \ type} \\\\maximum \ number= \frac{600,000 \ cm^2}{10,000 \ cm^2} \\\\maximum \ number= 60$

When all the three different shape types are used we can get the smallest number of tiles needed.

The minimum or smallest number of tiles needed (this will be possible if all the 3 different shapes are used)

$3 \times \ smallest \ number = 60\\\\smallest \ number = \frac{60}{3} \\\\smallest \ number = 20$

Thus, the smallest number of tiles Quintin will need in order to tile his floor is 20

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2 years ago

The largest set of x values satisfying

velikii [3]

Answer:

If $p\ge -\dfrac{1}{3},$ then $x>\dfrac{2}{3}+p$ and $mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p$

Step-by-step explanation:

Solve two inequalities for x.

1. $2,018x-p$

Separate terms with x and without x into two sides:

$2,018x-2,020x$

Multiply by -1:

$2x>-2p\\ \\x>-p$

2. $7x+3p$

Separate terms with x and without x into two sides:

$7x-10x$

Multiply by -1:

$3x>2+3p\\ \\x>\dfrac{2}{3}+p$

Find the largest set of x values satisfying both inequalities:

$x>-p\\ \\x>\dfrac{2}{3}+p$

$-p>\dfrac{2}{3}+p\\ \\-2p>\dfrac{2}{3}\\ \\2p$

then $x>-p$ and $mn=-p=p\cdot (-1),\ m+n=p-1.$ In this case both m and n are negative.

If $p>-\dfrac{1}{3},$ then $x>\dfrac{2}{3}+p$ and $mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p$

If $p=-\dfrac{1}{3},$ then $x>\dfrac{1}{3}$ and $mn=\dfrac{1}{3}=\dfrac{1}{3}\cdot 1,\ \ m+n=\dfrac{4}{3}$

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2 years ago