All categories

3 years ago

What is the highest common factor 12 and 30 what does factor mean

2 answers:

6 0

A common factor is a number that can be divided in two numbers. 6 can be divided into 12. 6*2=12
And 6 can also be divided into 30 too. 6*5=30

So the factor of 12 and 30 is 6.
I hope this helps!!
Let me know!
+ give me brainliest answer please!!

lions [1.4K]3 years ago

5 0

6. A common factor is a number that can be divided in both numbers, so 6 can go into 12 and 30, so it is a common factor

You might be interested in

3k - 2c + 8<br> use k = 7 and c = 4

Inga [223]

Answer:

Step-by-step explanation:

3 (7) - 2(4) + 8

21 - 8 + 8

13 + 8

6 0

3 years ago

Read 2 more answers

What is the prudct of 7\9 and 6

mezya [45]

Answer:

378

Step-by-step explanation:

9 × 7 × 6 = 378

pls mark brainliest, if not, its ok :)

6 0

3 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

The surface area of this prism is _______ cm squared

Katen [24]

Can you help me with my posted questions

6 0

3 years ago

I need help what option

uysha [10]

Answer:

24 es la b espero que te sirva

7 0

3 years ago