$6\sqrt3\cdot\dfrac{\sqrt3}{3}=\dfrac{(6\sqrt3)(\sqrt3)}{3}\\\\\text{use the associative property}\ (a\times b)\times c=a\times(b\times c)\\\\=\dfrac{(6)(\sqrt3\cdot\sqrt3)}{3}\\\\\text{use}\ \sqrt{a}\cdot\sqrt{a}=a\\\\=\dfrac{(6)(\not3)}{\not3}\qquad\text{cancel 3}\\\\=6$

3 0

2 years ago

Read 2 more answers

Tom plants 3 seeds.

Rasek [7]

Answer:

A. 64/125

B. 124/125.

Step-by-step explanation:

A). As the events ( germinate or not germinate) are independent we multiply the probabilities.

Prob(All seeds germinate) = 4/5*4/5*4/5 = 64/125.

B). Probability of at least one germinating = 1 - probability that none germinate

Probability of 1 seed not germinating = 1 -45 = 1/5.

So Prob(at least one germinating)

= 1 - (1/5 * 1/5 * 1/5)

= 1 - 1/125

= 124/125.

5 0

2 years ago

If f(x)=3x-5 and g(x)=x+3,find (f-g) (x)<br> A.2x-2<br> B.8-2x<br> C.4x-2<br> D.2x-8

son4ous [18]

(f-g)(x)=(3x-5)-(x+3) = 3x-5-x-3 = 2x-8
Therefore the answer is D.

Hope this helps.

5 0

3 years ago