All categories

3 years ago

Can you divide out common factors

1 answer:

5 0

Well you just divide fractions, invert the divisor and or second fraction and multiply. Then simplify by canselling out the common factors..... multiply and simplify.
Hope this helps~

You might be interested in

If 18 is added to a certain number and the sum is divided by 6, the result is 1 more than 14 of the original number. Find the or

Katyanochek1 [597]

The original number is 72

(18+x)/6 = 1+14
1/6x + 3 = 1 + 14 (Distribute)
1/6x + 3 = (1+14) (Combine Like Terms)
1/6x + 3 = 15
- 3 = -3 (Subtract 3 From Both Sides)
1/6x = 12
(1/6x)*6 = 12 * 6 (Multiply Both Sides By 6)
x = 72

5 0

3 years ago

How to round to nearest hundredth

loris [4]

Answer:

just look at the decimal after the decimal number is the tenths place the hundreds place and so on if i have to round 567.1092 i would round the nine up to make the zero a one so my rounded answer would be 567.11

Step-by-step explanation:

4 0

2 years ago

James orders clothing from an online retailer. James spends $105.91 on three pair of jeans, after using a $20 gift card. If the

Hoochie [10]

Answer: 84.728

Step-by-step explanation: 105.91•.20=21.182

105.91-21.182=84.728

If you have any questions just ask

7 0

3 years ago

Find the unit rate<br> 180 miles in 4 hours =<br> miles per hour

Andrei [34K]

Answer:

= 45 miles per hour

Step-by-step explanation:

First, we want a unit rate where 1 is in the denominator.

So we divide,

180 miles/ 4 hours both ÷ by 4

= 45 miles/ 1 hour

= 45 miles/ hour

= 45 miles per hour

5 0

2 years ago

Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f

nata0808 [166]

Via Lagrange multipliers:

$L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)$
$L_x=2xy+1+\lambda y=0$
$L_y=x^2+1+\lambda x=0$
$L_\lambda=xy-5=0$

$\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y$
$xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x$

$\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}$
$xy=5\implies y=\pm\sqrt{30}$

At these points, we get local minima of $f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}$ .

- - -

Another way to do this is to make $f(x,y)$ a function independent of $y$ , which is made possible by the constraint.

$xy=5\implies y=\dfrac5x$
$\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x$
$\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}$

and so on.

8 0

3 years ago