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3 years ago

Can someone help me on this please

Mathematics

2 answers:

a_sh-v [17]3 years ago

5 0

Answer:

Step-by-step explanation:

Lady bird [3.3K]3 years ago

3 0

The answer would be 3, I hope you have a nice day

May I have brainly? It’s okay if not ..:)

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A point with coordinates (a, b) is plotted on a coordinated plane. The values of a and b can be any positive or negative integer

Diano4ka-milaya [45]

Answer:

Any negative and positive number can be your coordinates.

Step-by-step explanation:

As long as you don't have any restrictions as to what the coordinates can be, you're good to go!

Hope this helps!

8 0

3 years ago

When you open a stepladder, you use a brace on each side of the ladder to lock the legs in place. Explain why the triangles form

navik [9.2K]

Answer:

so it can have a better balence

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Jenny's bakery sells carrot muffins at 2 dollars each. The electricity to run the over is over 120 dollars per day and the cost

g100num [7]

Answer:

200 muffins

Step-by-step explanation:

The selling price of each muffin = $2

The cost price of each muffin = $1.40

The price of electricity = $120

If the number of muffins that Jenny sells in a day is x, then, Jenny's total cost in a day is:

120 + (1.4 * x) = 120 + 1.4x

and the total sales earnings in a day for x muffins will be:

2 * x = 2x

To break even, the total costs in a day must equal to the total earnings. That is;

$2x = 120 + 1.4x$

Solving this:

$2x - 1.4x = 120\\\\\\0.6x = 120\\\\\\x = 120/0.6\\\\\\x = 200$

She must sell 200 muffins in a day to break even.

7 0

3 years ago

HELP PLEASE ASAP<br> I will mark brainliest if you explain

scoray [572]

Answer:

y - 5, (x, y) - (0,5)

3 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

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

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago