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Sunny_sXe [5.5K]
3 years ago
5

Please show me 3+2 (2x - 4) and can you please show the work?

Mathematics
2 answers:
Julli [10]3 years ago
7 0

Answer:

4x -5

Step-by-step explanation:

3+2 (2x - 4)

Distribute

3 + 2*2x - 2*4

3+4x - 8

Combine like terms

4x +3-8

4x -5

wlad13 [49]3 years ago
7 0

Answer:

4x - 5

Step-by-step explanation:

3+ 2 (2x) + 2 (-4)

3+4x - 8

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3( x -5)- 5x= 25+ 6x What is the answer and how do you solve this??
stiks02 [169]
1.) expand

3x - 15 - 5x = 25 + 6x

2.) simplify

-2x - 15 = 25 + 6x

3.) add 2x to both sides

-15 = 25 + 6x + 2x

4.) simplify

-15 = 25 + 8x

5.) subtract 25 from both sides

-15 - 25 = 8x

6.) simplify -15 - 25 to -40

-40 = 8x

7.) divide both sides by 8

-40/8 = x

8.) x = -5


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3 years ago
The weights of a certain brand of candies are normally distributed with a mean weight of 0.8593 g and a standard deviation of 0.
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6 0
3 years ago
Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)
Paladinen [302]

Answer:

\frac{140}{3}, \frac{140}{3}, \frac{140}{3}

Step-by-step explanation:

Let the numbers be x, y\ and\ z.

Such that:

x + y + z = 140

Make z the subject

z = 140 -x - y

For their product to be maximum, we have:

f(x,y,z) = xyz

Substitute z = 140 -x - y in f(x,y,z) = xyz

f(x,y) = xy(140 - x - y)

Open bracket

f(x,y) = 140xy - x^2y - xy^2

Differentiate w.r.t x and y

f_x=140y - 2xy - y^2

f_y=140x - x^2 - 2xy

Since the products are maximum, then f_x = f_y = 0

For f_x=140y - 2xy - y^2

140y - 2xy - y^2 = 0

Factorize:

y(140 - 2x - y) = 0

Split

y = 0\ or\ 140 - 2x - y = 0

Make y the subject

y = 0\ or\ y = 140 - 2x

For f_y=140x - x^2 - 2xy

140x - x^2 - 2xy = 0

---------------------------------------------------

Substitute y = 0

140x - x^2 -2x*0 = 0

140x - x^2 = 0

Factorize

x(140 - x)= 0

x = 0\ or\ 140-x = 0

x = 0\ or\ x = 140

---------------------------------------------------

Substitute y = 140 - 2x

140x - x^2 - 2xy = 0

140x - x^2 - 2x(140 - 2x) = 0

140x - x^2 - 280x + 4x^2 = 0

Re-arrange

4x^2 -x^2 +140x - 280x = 0

3x^2 -140x = 0

Factor x out

x(3x - 140) = 0

Divide through by x

3x - 140 = 0

3x = 140

x = \frac{140}{3}

Recall that: y = 140 - 2x

y = 140 - 2 * \frac{140}{3}

y = 140 - \frac{280}{3}

Take LCM

y = \frac{140*3-280}{3}

y = \frac{140}{3}

Recall that:

z = 140 -x - y

z = 140 - \frac{140}{3} - \frac{140}{3}

Take LCM

z =  \frac{3 * 140- 140 - 140}{3}

z =  \frac{140}{3}

Hence, the numbers are:

\frac{140}{3}, \frac{140}{3}, \frac{140}{3}

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Answer:

answer: obtuse

Step-by-step explanation:

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