All categories

3 years ago

-3x-(3x-1)=-4x-9 what is the answer

Mathematics

1 answer:

Karolina [17]3 years ago

3 0

Answer: x = 5

Step-by-step explanation:

Want me to show this, or did you just want the answer?

You might be interested in

5 Ib equals how many oz

Snezhnost [94]

5 lbs equals 80 ounces because 1 lb is 16 oz

3 0

4 years ago

Read 2 more answers

Find the range of the following set of data:12,2,14,80,100

Leviafan [203]

Answer: 98

Step-by-step explanation: Subtract the biggest number from the smallest. 100-2=98

8 0

3 years ago

Read 2 more answers

Someone help pls i will mark u brainliest

Alla [95]

Answer:

Step-by-step explanation:

cant see the picture

8 0

3 years ago

Read 2 more answers

What is the HCF of 60 and 96?

pochemuha

Answer:
12

Explanation:
20 words do be annoying

4 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

-3x-(3x-1)=-4x-9 what is the answer​

-3x-(3x-1)=-4x-9 what is the answer