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Marianna [84]
3 years ago
9

Where do the parentheses go and solve 3×14=3×2×7=3×2×7=×7

Mathematics
1 answer:
grin007 [14]3 years ago
8 0

Answer:

I  don't think this is a real math equation.

Step-by-step explanation:

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Find the median of the following data:
Nezavi [6.7K]

Answer:

c is the answer

Step-by-step explanation:

4 0
4 years ago
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ohan found that the equation –2|8 – x| – 6 = –12 had two possible solutions: x = 5 and x = –11. which explains whether his solut
klio [65]
-2 | 8-x| - 6 = -12...add 6 to both sides
-2 |8-x| = -12 + 6
-2 |8-x| = - 6...divide both sides by -2
|8-x| = -6/-2
|8-x| = 3

8 - x = -3          -(8 - x) = -3
-x = -3-8           -8 + x = -3
-x = -11              x = -3 + 8
x = 11                x = 5

not correct.....sign error
5 0
4 years ago
Write out the sample space for the given experiment. Use the following letters to indicate each choice: W for white, Y for yello
allochka39001 [22]

Answer:

S={WG, WE, WM, YG, YE, YM, BG, BE, BM}

Step-by-step explanation:

There are three possible color choices for the game room (White, Yellow or Blue), and three possible choices for the entertainment center (Gray, Ebony or Maple). This means that there are 9 possible outcomes, therefore, the sample space for the experiment is:

S={WG, WE, WM, YG, YE, YM, BG, BE, BM}

4 0
3 years ago
PLZZ HELP I NEED HELP
Mkey [24]

Step-by-step explanation:

1) let the number=x

six times a number=6x

Condition:

6x+4=22

2) eleven times a number=11x

Condition:

11x-5=50

3) 9 times a number=9x

Condition:

9x-7=-16

<u>N</u><u>o</u><u>t</u><u>e</u><u>:</u><u>i</u><u>f</u><u> </u><u>y</u><u>o</u><u>u</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>a</u><u>s</u><u>k</u><u> </u><u>a</u><u>n</u><u>y</u><u> </u><u>question</u><u> </u><u>please</u><u> </u><u>let</u><u> </u><u>me</u><u> </u><u>know</u><u>.</u>

3 0
3 years ago
CALCULUS: Determine which function is a solution to the differential equation y ' − y = 0.
Montano1993 [528]

C: none of these are solutions to the given equation.

• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.

• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.

The actual solution is easy to find, since this equation is separable.

<em>y'</em> - <em>y</em> = 0

d<em>y</em>/d<em>x</em> = <em>y</em>

d<em>y</em>/<em>y</em> = d<em>x</em>

∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>

ln|<em>y</em>| = <em>x</em> + <em>C</em>

<em>y</em> = exp(<em>x</em> + <em>C </em>)

<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>

8 0
3 years ago
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