Ask question

Ask question

All categories

3 years ago

15

Need help with this problem

2 answers:

lesantik [10]3 years ago

4 0

2sinxcosx is equal to sin(2x)

therefore,

= sin(2x)/cos(2x)
= tan(2x) (ANSWER)

pickupchik [31]3 years ago

3 0

2sinxcosx = sin2x
:. Sin2x / cos2x = tan2x = 2tanx / 1-tansquarex

You might be interested in

PLEASE HELP ANYONE

LUCKY_DIMON [66]

For this problem all you have to do is plug in the value that they is in parenthesis for x. If it says g (x) = x and then it asks g(5) = ?, it is saying what happens if i put 5 in for every x. in this case it would be g (5) = 5. I just replaced x with 5.

So g (-2) we sub -2 for x
g (-2) = -2 (-2)^2 + 3 (-2) - 5
= -2 (4) - 6 - 5
= -19

g (0) = -2 (0) + 3 (0) - 5
= 0 + 0 - t
= -5

g (3) = -2 (3)^2 + 3 (3) - 5
= -18 + 9 - 5
= -14

6 0

4 years ago

A chess club has 500 members nut is losing 30 members each year. A computer club has 150 members and is gaining 20 members per y

Fiesta28 [93]

$500 - 30x = 150 + 20x$
$x = 7$

3 0

4 years ago

The first line of the passage includes a

Flauer [41]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A taxi ride costs p40.00 for the first 500 meters and each additional 300 meters or a fraction there of adds p3.50 to the fare.u

Zarrin [17]

The function which represents the taxi fare in terms of distance d in meter is f(d) = 3.50 × $\frac{d-500}{300}$ + 40 .

According to the question,

For first 500 meters cost is 40.00 and For each additional 300 meters cost is 3.50 .

Let, the total distance of taxi ride = d meters

Now, representing the equation of taxi fare in terms of distance when distance is less than 500

taxi fare = 3.50 × $\frac{d-500}{300}$ + 40

∴ f(d) = 3.50 × $\frac{d-500}{300}$ + 40

To know more about functions refer below link: brainly.com/question/12431044

#SPJ4

8 0

2 years ago

PLEASE PLEASE HELP

marta [7]

Answer:

Approximately 3 grams left.

Step-by-step explanation:

We will utilize the standard form of an exponential function, given by:

$f(t)=a(r)^t$

In the case of half-life, our rate <em>r</em> will be 1/2. This is because 1/2 or 50% will be left after <em>t </em>half-lives.

Our initial amount <em>a </em>is 185 grams.

So, by substitution, we have:

$\displaystyle f(t)=185\big(\frac{1}{2}\big)^t$

Where <em>f(t)</em> denotes the amount of grams left after <em>t</em> half-lives.

We want to find the amount left after 6 half-lives. Therefore, <em>t </em>= 6. Then using our function, we acquire:

$\displaystyle f(6)=185\big(\frac{1}{2}\big)^6$

Evaluate:

$\displaystyle f(6)=185\big(\frac{1}{64}\big)\approx2.89\approx 3$

So, after six half-lives, there will be approximately 3 grams left.

3 0

3 years ago