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3 years ago

A golfers score after playing on friday was +2

Mathematics

1 answer:

jek_recluse [69]3 years ago

6 0

The end of his round on Saturday was a -5, and the end of his round on Sunday was a 0.

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I) Find the first 4 terms in the expansion of

Katyanochek1 [597]

Answer:

(i) $64+192 x^2+240 x^4+160x^6$

(ii) 1072

Step-by-step explanation:

Part (i)

Using Binomial series formula:

$(2+x^2)^6=2^6+6C1 \cdot 2^{6-1}\cdot x^2+6C2 \cdot 2^{6-2}\cdot(x^2)^2+6C3 \cdot 2^{6-3}\cdot(x^2)^3$

$=64+6 \cdot 32\cdot x^2+15 \cdot 16\cdot x^4+20 \cdot 8 \cdot x^6$

$=64+192 x^2+240 x^4+160x^6$

Part (ii)

$(1-\frac{3}{x^2})^2=(1-\frac{3}{x^2})(1-\frac{3}{x^2})$

$=1-\dfrac{6}{x^2}+\dfrac{9}{x^4}$

$(2+x^2)^6(1-\frac{3}{x^2})^2=(64+192 x^2+240 x^4+160x^6)\left(1-\dfrac{6}{x^2}+\dfrac{9}{x^4} \right)$

$=64-\dfrac{384}{x^2}+\dfrac{576}{x^4}+192x^2-1152+\dfrac{1728}{x^2}+240x^4-1440x^2+2160+...$

There is no need to keep expanding since the remaining will include the variable.

Therefore, the term independent of x = 64 - 1152 + 2160 = 1072

6 0

2 years ago

Perform the operation and reduce the answer fully. 1/2 / 7/6

Shalnov [3]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A kite with a string 50 meters long makes an angle of elevation of 58o with the ground, assuming the string is straight, how hig

Sunny_sXe [5.5K]

An angle of elevation of 58 degrees with the ground means that the angle started at the x-axis and terminates on the first quadrant region through a counter clockwise direction. Imagining a triangle, the string of the kite becomes the hypotenuse, the angle 58 degrees is the angle between the hypotenuse and the ground, and the height of the kite is the side opposite to the angle.

Therefore, we can calculate for the height of the kite using the sin function:

sin θ = opposite side / hypotenuse

sin 58 = opposite side / 50 m

opposite side = 50 m * sin 58

opposite side = 42.4 m

Therefore the kite is about 42.4 m above the ground.

8 0

3 years ago

How to simplify -64^(1/3)

Nadusha1986 [10]