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3 years ago

12-3m=4+5/m HELPPPP

2 answers:

6 0

M1= 5 , M2=1

Explanation:

12-3m-4-5/m=0
8-3m-5/m=0
(8-3m-5/m=0) * m
8m-3m^2-5=0
8m-3m^2=5
m(8-3m)=5
m1=5
m2=> (8-3m)=5
3m=3
m2= 1

VladimirAG [237]3 years ago

4 0

Answer:

$m=1\\m=\frac{5}{3}$

Step-by-step explanation:

$12 - 3m = 4 + \frac{5}{m}$

$8=\frac{5}{m} +3m$

Multiply the whole thing by m

$8m = 5 +3m^{2}$

$3m^{2} -8m + 5 = 0$

$(3m^{2} -3m) + (-5m +5)$

$3m(m-1)-5(m-1)=0$

$3m-5=1\\m-1=0$

$m=1\\m=\frac{5}{3}$

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A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic

Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

$\mathbf{Bars = 6}$

$\mathbf{Chocolate = 3}$

$\mathbf{Custard= 3}$

The total number of donuts in the box is:

$\mathbf{Total= 6 + 3 + 3}$

$\mathbf{Total= 12}$

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

$\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}$

So, we have:

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}$

Simplify

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}$

Multiply

$\mathbf{Pr = \frac{9}{132}}$

Divide

$\mathbf{Pr = 0.068}$

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068