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Flauer [41]
3 years ago
6

What is the median number of excused absences?

Mathematics
1 answer:
zalisa [80]3 years ago
3 0

Answer:

3

Step-by-step explanation:

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Solve the following absolute value equation.
dolphi86 [110]

Hey there!

|\displaystyle\frac{x-1}{2} |=7

Let's say that x-1 is negative:

\displaystyle\frac{-x+1}{2} =7

The first step is to multiply both sides by 2:

\text{-x+1=14}

Now, move 1 to the right, using the opposite operation:

\text{-x=14-1}

\text{-x=13}

Divide both sides by -1:

\text{x=-13}

  • Note: Since the negative sign is already entered, all you need to enter is 13.

I hope you find my answer helpful.

Good luck!

Have a great day!

\bf{StargazingWithJoy}

5 0
3 years ago
Please Help! Will mark brainiliest!!<br><br> 1. y varies directly as x and inversely as z.
Paladinen [302]

Answer:

x = 5

z =2

y =4.4

Step-by-step explanation:

y varies directly as x

y = kx

and inversely as z

y = kx/z

Putting in the numbers in the first line of the table, we can solve for k

13.75 = k * 25/4

Multiply by 4

13.75*4 = 25k

55 =25k

Divide by 25

55/25 = 25k/25

2.2 = k

Now we can fill in the table

y = 2.2 x/z

Row 2 we need to find x

1 = 2.2 x/11

Multiply by 11

11 = 2.2 x

Divide by 2.2

11/2.2 = 2.2x

5 =x

Row 3 we need to find z

18.7 = 2.2 *17/z

18.7 = 37.4 /z

Multiply by z

18.7z = 37.4

Divide by 18.7

18.7/18.7z = 37.4/18.8

z =2

Row 4 we need to find y

y = 2.2 *10/5

y = 22/5

y =4.4

8 0
4 years ago
I keep getting confused I don't understand what I keep doing wrong
rosijanka [135]
\bf \begin{cases}&#10;5x+3y+z=-14\\&#10;2x-2y-z=-13\\&#10;3x+y+3z=-2&#10;\end{cases}

so.. let's us eliminate, say "z" first, so for that, let's pick the 1s and 2nd equations there  \bf \begin{array}{llll}&#10;5x+3y+z=-14\\&#10;2x-2y-z=-13\\&#10;----------\\&#10;\boxed{7x+y+0\ =-27}&#10;\end{array}

so, that's our first two-variables resultant

let's pick other two, and again, eliminate the same "z" variable, this time, let's use the 2nd and 3rd equations   \bf \begin{array}{llll}&#10;2x-2y-z=-13&\times 3\implies &6x-6y-3z=-39\\&#10;3x+y+3z=-2\implies &&3x+y+3z\ =-2\\&#10;&&----------\\&#10;&&\boxed{9x-5y+0\ =-41}&#10;\end{array}

now, we have two two-variables equations, let's use them then
and say, we'll eliminate the "y" variable from them

\bf \begin{array}{llll}&#10;7x+y=-27& \times 5\implies &35x+5y=-135\\&#10;9x-5y=-41\implies &&9x-5y=-41\\&#10;&&-------\\&#10;&&44x+0=-176&#10;\end{array}&#10;\\\\\\&#10;x=\cfrac{-176}{44}\implies \boxed{x=-4}&#10;\\\\\\&#10;\textit{now, that we know x = -4, let's plug that in }7x+y=-27&#10;\\\\\\&#10;7(-4)+y=-27\implies y=-27+28\implies \boxed{y=1}

\bf \textit{now, let's plug those two at }5x+3y+z=-14&#10;\\\\\\&#10;5(-4)+3(1)+z=-14\implies -20+3+z=-14\implies z=-14+17&#10;\\\\\\&#10;\boxed{z=3}
7 0
3 years ago
Identify the relationship of each angle pairs. Choose your answer from the choices given.
vova2212 [387]

Answer:

scheu7rggi6rewhjvzdeguij

5 0
3 years ago
What are the roots of the polynomial equation x4+x3=4x2+4x? Use a graphing calculator and a system of equations.
ivann1987 [24]
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Answer A.
6 0
4 years ago
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