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Blababa [14]
3 years ago
13

Who can solve this question​

Mathematics
1 answer:
Leviafan [203]3 years ago
3 0

Answer:n

ihn\ngbf x

x

vStep-by-step explanation: jjkd na

no[ncvfos

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Is the answer yes or no tell me please ?
diamong [38]

Answer:

Yes.

Step-by-step explanation:

4 0
3 years ago
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How can i find the area of this triangle?
Ivanshal [37]
To find the area of a triangle, multiply the base by the height, and then divide by 2. The division by 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, in the diagram to the left, the area of each triangle is equal to one-half the area of the parallelogram.
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Determine in which quadrant the product of 5(cos3pi/8 + isin 3pi/8) and sqrt 2(cos pi/12 + sin pi/12) lies.
ohaa [14]
When you multiply two complex numbers given in polar form, the argument of the product is the sum of the arguments of the factors. Meanwhile, the modulus of the product is the product of the moduli of the factors.


In this case, you'd have

\dfrac{3\pi}8+\dfrac\pi{12}=\dfrac{11\pi}{24}

and the modulus would simply be 5\sqrt2. Since

\dfrac{11\pi}{24}

we would expect the final product to fall in the first quadrant.
6 0
3 years ago
Please can anyone help me please​
Step2247 [10]
4.5, 4, 3.5, 3, 2.5, 2 and use the pencil and ruler, love a bit of MathsWatch
7 0
3 years ago
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Choose the appropriate pattern and use it to find the product: (p4−q4)(p4+q4).
sp2606 [1]

The expression can be solved by expanding the bracket and multiplying out the terms

(p^4-q^4)(p^4+q^4)\begin{gathered} =p^4(p^4+q^4)-q^4(p^4+q^4) \\ =p^8+p^4q^4-p^4q^4-q^8 \\ =p^8-q^8 \end{gathered}

Therefore, the expression can be simplified as;

p^8-q^8

Alternatively, using the theorem of difference of two squares, which is

a^2-b^2=(a-b)(a+b)

Hence,

p^8-q^8=(p^4)^2-(q^4)^2

7 0
1 year ago
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