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ExtremeBDS [4]
2 years ago
5

HELP PLEASE

Mathematics
2 answers:
Fiesta28 [93]2 years ago
8 0
The correct point of the graph would be B. (2,-5)
Harman [31]2 years ago
5 0

Answer:(2,-5)

Step-by-step explanation:

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Solve for the x in the inequality 6(2−6x)≤4−35x 6 ( 2 − 6 x ) ≤ 4 − 35 x .
dolphi86 [110]

Answer:

8 ≤ x

Step-by-step explanation:

6(2 − 6x) ≤ 4 − 35x

Expand bracket;

12 - 36x ≤ 4 − 35x

Add 36x to both sides to get;

12 ≤ 4 + x

Subtract 4 from both sides to get;

8 ≤ x

5 0
2 years ago
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PLEASE HELP WITH THIS ONE QUESTION
Nana76 [90]

Answer:

I'm not sure about the factors, but it's reflected over the x axis if they helps narrow it down

3 0
2 years ago
Write 7/25<br> as a terminating decimal.
ycow [4]

Answer: 0.28

Step-by-step explanation:

6 0
3 years ago
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Solve using algebraic equation: <br> 5sin2x=3cosx<br><br> (No exponents)
DaniilM [7]
5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0

Now the zero product property tells us that there are two cases where this is true,

\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of \dfrac\pi2, so x=\dfrac{(2n+1)\pi}2 where n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}

which occurs twice in the interval [0,2\pi) for x=\arcsin\dfrac3{10} and x=\pi-\arcsin\dfrac3{10}. More generally, if you think of x as a point on the unit circle, this occurs whenever x also completes a full revolution about the origin. This means for any integer n, the general solution in this case would be x=\arcsin\dfrac3{10}+2n\pi and x=\pi-\arcsin\dfrac3{10}+2n\pi.
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3 years ago
How to divide a given ratio into a given amount
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JfpeiJKFHGIFWjfowednklsv WEOIHGVFOEWDBISFKCJPFGVUOEIDSVB PIFGHFIAGHRKFNBLKGVBJ;ALKFBGO'klVFBFJSKDGPV;KFJDZNBAJGIJGJNVL;ZGJEORITUJLDNVZ;.JDFG[OAEJRGNA'JGBj'gaklrfmgo'ajgb'pzang'ajrebgmkza'pgjozn;klfdbgkl'fznkp'fgnko'lgkfv.

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