Ask question

Ask question

All categories

3 years ago

13

Both sherry and della work at a fast-food outlet. Sherry has every sixth day off and della has every forth day off. If they are

both off on Wdenesday of this week, what will be the day of the week that they are next off together? Show the work

1 answer:

erastovalidia [21]3 years ago

8 0

Answer:

Monday

Step-by-step explanation:

You might be interested in

There are 20 blue marbles and 10 red marbles in a bag .if a marble is drawn at random once ,what is the probability that it will

andrew11 [14]

Answer:

2/3

Step-by-step explanation:

First, you figure out how many marbles are in total, which is the denominator. The total number of marbles is 30.

Then, you figure out how many marbles are blue, which is the numerator. There are 20 blue marbles.

Finally, you simplify the fraction. 20/30=2/3.

5 0

3 years ago

Read 2 more answers

Find the greatest common factor of these two expressions 18y^6w^2 and 27y^5 w^4 u^7

drek231 [11]

Answer:

9 $y^{5}$ $w^{2}$

Step-by-step explanation:

Find what both expressions have in common

3 0

2 years ago

What is x if the other angles are 61 and 42

Lady_Fox [76]

Answer:

The answer is 180°. Hope this helps!

4 0

3 years ago

What is it 7 points on the board for u plz answer CORRECTTT

pochemuha

Answer:

x is 30

Step-by-step explanation:

1/2 ( x + 6 ) = 18

1/2x + 3 = 18

- 3 - 3

1/2x = 15

· 2 · 2

x = 30

5 0

2 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago