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3 years ago

Help I’ll mark you brainly!!

Mathematics

1 answer:

tester [92]3 years ago

6 0

the numbers to the questions are on the top left corner of each bubble, hope this helps!

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Hepl im desperate beacuse othis

Feliz [49]

You would have to do length times width. So you would do 6*6 then you will get 36. After that you would want to do 36*3. You would then get 108. The answer is B. 108 ft^2.

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3 years ago

What is the equation of the line that is perpendicular to and

sergey [27]

The 3rd one, Y=5x+1

5 0

3 years ago

Read 2 more answers

Lee fifth grade class has 1.4 times as many students as Fred's. If there are a total of 60 students. How many students does Fred

Kipish [7]

Lees class has 1.4 times more..
25 x 1.4= 35....which is Lee's total
Fred's would be the 25.
25 + 35 = 60
I just picked #s to start
example: 23x 1.4=32.2. 23+32.2=55.2...not 60

8 0

3 years ago

The rates of currency exchange published in the newspaper on a certain day showed that 12 pounds could be exchanged for 96 dolla

stealth61 [152]

Answer:

Step-by-step explanation:

Your answer

816 dollars

Mark it as Brainlist answer. Follow me.

7 0

3 years ago

Find the expansion of tan x about the point X = 0.

yan [13]

Answer:

$f(x) = x +\frac{1}{3}x^{3}+....$

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that <u>The Maclaurin series is given by:</u>

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

= 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

= 4sec²(x) tan²(x) + 2sec⁴(x)

= 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) = sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

$f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......$

$f(x) = x +\frac{1}{3}x^{3}+....$

Therefore, the expansion of tan x at x = 0 is

$f(x) = x +\frac{1}{3}x^{3}+....$ .

8 0

4 years ago