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3 years ago

7

How to get the number 4 with the numbers 2,3,6,8 using all of them and you only use it once.

1 answer:

sergey [27]3 years ago

5 0

Answer:

2 x 3 + 6 - 8

Step-by-step explanation:

2 x 3 = 6

6 + 6 = 12

12 - 8 = 4

:)

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A library had 6422 music CDs stored on 26 shelves. If the same number of CDs were stored on each shelf, how many CDs were stored

sattari [20]

Simply divide the number of CDs by the number of shelves. 6422 / 26 = 247 247 CDs per shelf.

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3 years ago

Read 2 more answers

Can you help me please.

Kamila [148]

9514 1404 393

Answer:

a. (-1/2, 5)

Step-by-step explanation:

The coefficients of x have opposite signs, so the x-terms can be eliminated by adding the two equations.

(2y -12x) +(3y +12x) = (16) +(9)

5y = 25 . . . . . . simplify

y = 5 . . . . . . . . .divide by 5

This matches only answer choice A.

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<em>Check</em>

The other part of answer choice A is x = -1/2. We can test this to make sure it works:

2(5) -12(-1/2) = 16

10 + 6 = 16 . . . . . . true, the selected answer works OK

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2 years ago

Smooth, suave, ingratiatingly pleasant

mash [69]

I say the answer is B. Bland

4 0

3 years ago

I know it is basic but I need help<br><br>11m+13=m+23

zloy xaker [14]

Answer:

m = 1

Step-by-step explanation:

11m + 13 = m + 23

<u>Get m by itself by subtracting m from both sides. </u>

10m + 13 = 23

<u>Subtract 13 from both sides.</u>

10m = 10

<u>DIvide by 10.</u>

m = 1

8 0

3 years ago

Read 2 more answers

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:

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

8 0

2 years ago