What two numbers whose sum is 31 and whose difference is 1

The Lopez family is selecting a furniture set. A furniture set has a bed, a desk and a dresser. There are 2 beds, 3 desks, and 2

Musya8 [376]

2x3x2 witch equals 12 so there are 12 different furniture sets

3 0

2 years ago

write the slope-intercept form for the equation of the line described through (4,-5) and parallel to y=--5/4x-4

Tems11 [23]

Answer:

y = -5x/4 + 0

Step-by-step explanation:

Find the slope

y = -5/4x - 4

Slope is the coefficient of x

Slope m = -5/4

Substitute m into point slope form equation

y - y1 = m(x - x1)

y - y1 = -5/4( x - x1)

Substitute the point into the equation

( 4 , -5)

x1 = 4

y1 = -5

y - y1 = -5/4( x - x1)

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

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

Using a slope intercept form equation

y = mx + c

y - intercept point y

m - slope

x - intercept point x

c - intercept

open the bracket with -5/4

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

y + 5 = (-5x + 20)/4

y = (-5x + 20)/4 - 5

LCM = 4

y =( -5x + 20 - 20)/4

y = ( -5x + 0)/4

Rearrange in

y = mx + c

y = -5x/ 4 + 0/4

y = -5x/4 + 0

The equation of the line is

y = -5x/4 + 0

3 0

2 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .