All categories

2 years ago

Multiply (3x-5)(x+4)

Mathematics

2 answers:

Allisa [31]2 years ago

4 0

Answer:

3x^2+7x-20

Step-by-step explanation:

77julia77 [94]2 years ago

3 0

Answer:

expand the brackets

3x^2 + 7x -20

Step-by-step explanation:

(3x-5)(x+4)

= 3x^2 + 12x - 5x + 20

= 3x^2 + 7x + 20

You might be interested in

Center -c (3,-2), radius 4.

taurus [48]

Answer:

(x-3)^2+(y+2)^2=16

Step-by-step explanation:

center (h,k) radius r

(x-h)^2+(y-k)^2=r^2

6 0

3 years ago

What is the value of y?

Colt1911 [192]

The answer is 14. The two lower angles are 50 each 100 total and so the top angle must equal 80. If you set it equal to 80 you get 14.

4 0

3 years ago

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

PLZZZZ HEEEEEEELLLPPP MEEEEE!

lesantik [10]

Answer:

144 it is D

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Please help!!

ivanzaharov [21]