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joja [24]
3 years ago
5

The diagram shows a sector of circle radius 10 cm find the area of the sector to one decimal place

Mathematics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

314 cm²

Step-by-step explanation:

Area of a circle is A= πr²

pi is 3.14 and radius is 10:

A= 3.14(10)²

A= 314

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5. 18 m 11 m Not drawn to scale 29 m b. 445 m 7 m d. 203 m​
Luden [163]
Answer: D

Explanation: Apply the pythagorean theorem (a^2 + b^2 = c^2) to get 11^2 + b^2 = 18^2. From that you can solve for b and you get the square root of 203.

Hope this helped! :)
8 0
2 years ago
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PLZZZZ ANSWER FAST At the beginning of spring, Robert planted a small sunflower in his backyard. When it was first planted, the
Mademuasel [1]

Answer:

um 35....... i think. sorry if my answer is wrong.

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2 years ago
Whats the slope of (-8,-1) and (-6,-2)
STatiana [176]
Answer: -1/2 or -0.5.
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2 years ago
A triangle with an area of 37 units? is dilated by a scale factor of . Find the area of
sergeinik [125]

Answer:

<u>58 units</u>

Step-by-step explanation:

I decided that one side is 37 units and another is 2 units, since that would make the area 37 units. (you can also use 74 units and 1 unit)

Then I multiplied 37 by 5/4, which equals a new length of 46.25 units, and I also multiplied 2 by 5/4, which equals a new length of 2.5 units.

Finally, I solve for the area of the triangle:

1/2(46.25 x 2.5) ≈ <u>58 units</u> (rounded to the nearest whole number)

7 0
2 years ago
20 points and brainliest <br> I’m in quiz in need it asap <br> Number 4
iren [92.7K]

Answer and step-by-step explanation:

The polar form of a complex number a+ib is the number re^{i\theta} where r = \sqrt{a^2+b^2} is called the modulus and \theta = tan^-^1 (\frac ba) is called the argument. You can switch back and forth between the two forms by either remembering the definitions or by graphing the number on Gauss plane. The advantage of using polar form is that when you multiply, divide or raise complex numbers in polar form you just multiply modules and add arguments.

(a) let's first calculate moduli and arguments

r_1 = \sqrt{(-2\sqrt3)^2+2^2}=\sqrt{12+4} = 4\\ \theta_1 = tan^-^1(\frac{2}{-2\sqrt3}) =-\pi/6\\r_2=\sqrt{1^2+1^2}=\sqrt2\\ \theta_2 = tan^-^1(\frac 11)= \pi/4

now we can write the two numbers as

z_1=4e^{-i\frac \pi6}; z_2=e^{i\frac\pi4}

(b) As noted above, the argument of the product is the sum of the arguments of the two numbers:

Arg(z_1\cdot z_2) = Arg(z_1)+Arg(z_2) = -\frac \pi6 + \frac \pi4 = \frac\pi{12}

(c) Similarly, when raising a complex number to any power, you raise the modulus to that power, and then multiply the argument for that value.

(z_1)^1^2=[4e^{-i\frac \pi6}]^1^2=4^1^2\cdot (e^{-i\frac \pi6})^1^2=2^2^4\cdot e^{-i(12)\frac\pi6}\\=2^2^4 e^{-i\cdot2\pi}=2^2^4

Now, in the last step I've used the fact that e^{i(2k\pi+x)} = e^i^x ; k\in \mathbb Z, or in other words, the complex exponential is periodic with 2\pi as a period, same as sine and cosine. You can further compute that power of two with the help of a calculator, it is around 16 million, or leave it as is.

7 0
1 year ago
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