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3 years ago

Quick help! it EASY

Mathematics

2 answers:

gulaghasi [49]3 years ago

8 0

Answer:

IT EASY

Step-by-step explanation:

Anton [14]3 years ago

7 0

Answer:

the answer is 38034

Step-by-step explanation:

(x+18)(x-18) or (-x+18)(-x-18)

2x squared+13x+20

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Find the area of the composite figure will give extra 100 points out if answer is right and brainlyst

Vera_Pavlovna [14]

Answer:

Decompose the figure into two rectangles and add the areas.

Find the area of the entire rectangle and of the removed corner and subtract the areas.

Decompose the figure into three rectangles and add the areas.

Step-by-step explanation:

With all of these you can actually calculate the area of the composite figure, some of them are more easy and efficient than the other, for example dividing the composite area into three rectangles is not very efficient but will do the job, and the one where you decompose the area into two rectangles would be the best one, as well as the one where you find the area of the larger rectangle and the subtract from that the rectangle that is taken off in the right corner.

5 0

2 years ago

Evaluate the limit of tan 4x/ 4tan3x

Brut [27]

Answer:

1/3

Step-by-step explanation:

The ratio is undefined at x=0, so we presume that's where we're interested in the limit. Both numerator and denominator are zero at x=0, so L'Hôpital's rule applies. According to that rule, we replace numerator and denominator with their respective derivatives.

$\displaystyle\lim\limits_{x\to 0}\dfrac{\tan{(4x)}}{4\tan{(3x)}}=\lim\limits_{x\to 0}\dfrac{\tan'{(4x)}}{4\tan'{(3x)}}=\lim\limits_{x\to 0}\dfrac{4\sec{(4x)^2}}{12\sec{(3x)^2}}=\dfrac{4}{12}\\\\\boxed{\lim\limits_{x\to 0}\dfrac{\tan{(4x)}}{4\tan{(3x)}}=\dfrac{1}{3}}$

6 0

2 years ago

You have a fishing line spool with an end that has an area of 20.0 cm2. how much fishing line do you need to wind around the spo

grandymaker [24]

Answer: 158.53 cm

Step-by-step explanation:

We know that the ends of a fishing line spool are circular in shape.

Given : The area of an end of fishing line spool = $20.0\cm^2$ (1)

Area of a circle = $\pi r^2$ (2)

Circumference of a circle = $2\pi r$ (3)

, where r is radius of the circle.

From (1) and (2), we have

$\pi r^2=20\\\\\Rightarrow\ r^2=\dfrac{20}{\pi}\\\\\Rightarrow\ r=\sqrt{\dfrac{20}{\pi}}$

Circumference of fishing spool = $2\pi r$ (using (3))

$=2\pi \sqrt{\dfrac{20}{\pi}}=2\sqrt{20\pi}=2\times2\sqrt{5\pi}=4\sqrt{5\pi}$

i.e. Fishing spool required to wind around the spool one time $=4\sqrt{5\pi}\ cm$

⇒ Fishing spool required to wind around the spool 10 times $=10\times4\sqrt{5\pi}\ cm=40\sqrt{5\pi}\\\\=40\times\sqrt{5}\times \pi\\\\=40\times2.236\times\sqrt{3.14159}\\\\=89.44\times1.772453\\\\=158.528205473\approx158.53\ cm$