All categories

2 years ago

Can someone please help

Mathematics

2 answers:

Alika [10]2 years ago

4 0

The first option is the one you should pick

Vika [28.1K]2 years ago

3 0

It the second one

The anwser is the second one

You might be interested in

The amount of money you earn varies directly to the number of pizzas you sell. If you earn $44 for selling 8 pizzas, How many pi

Anna [14]

Answer:

You would need to sell 24 pizzas

Step-by-step explanation

132/44= 3

Take the 3 and times it by 8

3 x 8 = 24

You would need to make 24 pizzas

Each pizza is 5.50$

So an equation for this could be

5.50x P = C

P = Pizza

and C is the cash you earned

I hope this helps

5 0

2 years ago

What are some ways you can create a new system of equations based on a given system?

docker41 [41]

By adding/ subtraction systems

3 0

3 years ago

What is 6 5/3 x 5 7/8 < 42

Eddi Din [679]

Answer:

The left side 45.041¯645.0416‾ is not less than the right side 42 which means that the given statement is false.

3 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

I dont understand this can someone give me the answer. "Determine the equation of the line represented on the graph below.

erma4kov [3.2K]

Answer:

y=2/5x-1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers