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8090 [49]
2 years ago
14

Find the HCF and LCM of: x^3+27,2x^3-6x^2+18x,x^2-3x+9​

Mathematics
1 answer:
r-ruslan [8.4K]2 years ago
7 0

Answer:

hcf:3x^3-6x^2+y^2-3x+54

lcm:2(x+3)(x^2-3x+9)(x^3-3x^2+9)(y^2-3x+9)

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(1+2i) (2+5i)
Lostsunrise [7]

Answer:

Step-by-step explanation:

first multiply 1 with 2+5i then multiply 2i with 2+5i and you will get

2+5i+4i+10i^2

then in the next step add 4i and 5i you will get 9i

in the next step put i^2=-1 and you will get -10

in the last step just substract 2-10 you will get -8 and 9i

and your answer will be -8+9i

5 0
3 years ago
Who was found the math​
Zolol [24]
<h3>Answer:</h3>

  • Math is not discovered, It's Invented.

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✍︎ꕥᴍᴀᴛʜᴅᴇᴍᴏɴǫᴜᴇᴇɴꕥ

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4 0
3 years ago
find the area of a circke with a radius of 9 mm. round to the nearest tenth. use 3.15 or 22/7 for Pi.​
GREYUIT [131]

Answer:

254.3 mm^2

Step-by-step explanation:

Area of a circle: A = π*r^2

π (pi) = 3.14

r (radius) = 9 mm

Substitute 3.14 for pi and 9 for radius:

A = π*r^2

A = 3.14*9^2

A = 3.14*81

A = 254.34 mm^2

Round to the nearest tenth:

254.34 ==> 254.3

Therefore, our final answer is 254.3 mm^2

Hope this helps!

6 0
2 years ago
What is the value of Cos n 2π=???
faust18 [17]
If n is an integer, then cos(n(2<span>π))=1.

Take n=0, cos(2</span>π)=cos(0)=1, n=1, cos(2π)=1, n=2, cos(2(2π))=cos(4<span>π)=cos(0)=1...

If n is not an integer, well, ..... we don't know the value, which depends on the value of n.</span>
5 0
3 years ago
On a unit circle, the ordered pair (x, y) represents the point where the terminal side of e intersects the unit circle. If m&lt;
tatyana61 [14]

Complete Question

Refer to attachment

Answer:

Option C

Step-by-step explanation:

From the question we are told that:

m \angle \theta=120

Generally the angle of inclination is \theta

\theta =180-120

\theta=60

Generally the equation for x is mathematically given by

x=cos60\textdegree *1

x=\frac{1}{2}

Since x exists in the second quadrant

Therefore the value of x in simplest  form is

 x=-\frac{1}{2}

Option C

4 0
3 years ago
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