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3 years ago

What is the absolute value for 24.5

Mathematics

2 answers:

miv72 [106K]3 years ago

5 0

Yes, 24.5 is the absolute value!

Hope this helps

Ilya [14]3 years ago

3 0

Its 24.5
when its absolute value, everything goes to positive.

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(-3, 1), (-17, 2) find the slope show the work

Verizon [17]

Answer:

1/-14 or -0.071428571428571

Step-by-step explanation

(-3,1) ___ -3 is x1 while 1 is y1

(-17, 2) ___ -17 is x2 while 2 is y2

slope formula is m = y2-y1/x2-x1

plug them in: m = 2-1/-17-(-3), which equals 1/-14 or -0.071428571428571

5 0

3 years ago

The first term of an arithmetic sequence is -30 and the common difference is 8. Find the 20th term.

Sedaia [141]

Answer:

122.

Step-by-step explanation:

nthe term = a1 + d(n - 1) so here:

20th term = -30 + 8(20 - 1)

- 30 + 152

= 122.

4 0

3 years ago

There are 312 cards in 6 deck of playing cards write the ratio as a fraction them find the unit rate

ladessa [460]

Answer:

6x: 312

x: 52

Step-by-step explanation:

Let the deck be denoted by x then

6x: 312

There are 312 cards in 6x decks

In 1 deck x there are 312/6= 52 cards

6x: 312

6x/6: 312/6 Cancelling both sides by 6

x: 52

In ratio both the sides are divided by the same number to get equivalent fractions.

In 1 deck there are 52 cards

3 0

3 years ago

3m/m-6 times 5m^3-30m^2/5m^2

S_A_V [24]

-6x^4-30x^3+3

hope it helps

5 0

3 years ago

PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%

Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

8 0

3 years ago