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fomenos
3 years ago
8

Double a number increased by Ten equals twelve express the solution in 3 different ways

Mathematics
1 answer:
Zina [86]3 years ago
7 0
First way: 1 doubled is 1+1, which is 2. 2 increased by 10 equals 12.

Second way: 1 doubled is 1*2, which equals 2. 2 increased by 10 equals 12.

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Write an equation in standard form of a line that goes through these two points (-1, 1) and (5, -3). Help, please!
MatroZZZ [7]

Step-by-step explanation:

assuming two points with coordinates :

(Xa,Ya) and (Xb,Yb)

use formula : (Xb-Xa)/ (Yb-Ya)

= 5+1/-3-1 = 6/-4

the standard form is 6/-4 × x

8 0
3 years ago
I need help with this question
Novay_Z [31]

Answer:

$ \frac{\sqrt{3} - 1}{2\sqrt{2}} $

$ \frac{-(\sqrt{3} + 1)}{2\sqrt{2}} $

$ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $

Step-by-step explanation:

Given $ \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} $

(A) $ sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4}  + \frac{\pi}{6}) $

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$ sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2} $

$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}

(B) Cos(A + B) = CosAcosB - SinASinB

$ cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}}) $

$ \implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} $

$ \implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}) $

$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}

(C) Tan(A + B) = $ \frac{Sin(A +B)}{Cos(A + B)} $

From the above obtained values this can be calculated and the value is $ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $.

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Find the area of the shape shown below.
tekilochka [14]

Answer:

about 23.25

Step-by-step explanation:

i did the math.

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