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valentina_108 [34]
3 years ago
7

WILL MARK BRANLIEST, FOLLOW YOU, AND SAY UR THE BEST ON UR PROFILE

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0
The first one should be -24
2 is d there is no reason for those numbers to be negative
3=B
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What is the solution set to the inequality
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(4x - 3)(2x - 1) \geqslant 0 \\ \Leftrightarrow \begin{cases}4x - 3 \geqslant 0 \\ 2x - 1 \geqslant 0\end{cases}\: \vee \:\begin{cases}4x - 3 \leqslant 0 \\2x - 1 \leqslant 0 \end{cases} \\ \Leftrightarrow \begin{cases}x \geqslant  \frac{3}{4}  \\ x  \geqslant  \frac{1}{2} \end{cases}\: \vee \:\begin{cases}x\leqslant  \frac{3}{4}  \\x \leqslant  \frac{1}{2}  \end{cases} \\ \Leftrightarrow x \geqslant  \frac{3}{4} \: \vee \: x \leqslant  \frac{1}{2}  \\ \Rightarrow The\: third\: option

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Step-by-step explanation:

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Solve in radians help plz
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I think we can use the identity  sin x/2  = sqrt [(1 - cos x) /2]

cos x -  sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1

cos x - sqrt(3/2) sqrt(1 - cos x) = 1
sqrt(3/2)(sqrt(1 - cos x) =  cos x - 1   Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1

cos^2 x - 0.5 cos x - 0.5 = 0 

cos x = 1 , -0.5

giving x = 0 , 2pi, 2pi/3,  4pi/3  ( for  0  =< x <= 2pi)

because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.

The last 2 results dont fit so the answer is  x = 0 , 2pi Answer
5 0
3 years ago
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