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4 years ago

16-2t=5t+9 any answers

Mathematics

2 answers:

kirza4 [7]4 years ago

7 0

Answer:

t = 1

Step-by-step explanation:

Combine the t terms on the right side, so that the resulting t term will have a positive coefficient:

16 = 7t + 9

Subtract 9 from both sides, to isolate 7t:

7 = 7t

Divide both sides by 7 to obtain a solution:

t = 7/7 = 1

t = 1

just olya [345]4 years ago

3 0

The answer is: t = 1

You need to separate t from everything to figure out what it is.

Step 1: Subtract 9 from both sides, so we get 7 - 2t = 5t

Step 2: Add 2t to both sides, so we get 7 = 7t.

Step 3: Divide by 7, meaning t = 1.

Next you should check your work. If we plug 1 in for t, we get 14 = 14, so the answer is correct.

Hope this helps!

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2 years ago

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3 years ago

Read 2 more answers

Please help math questions put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

Problem 1

$\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}$

Therefore, the second option is correct.

Problem 2

$cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)$

$0=sinx\\\\x=\pi n\\\\x=0,\pi$

$0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, the solution set is $x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}$ , making the second option correct.

Problem 3

$2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0$

$cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ$

$2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ$

Therefore, the solution set is $x=\bigr\{90^\circ,120^\circ,240^\circ,270^\circ\bigr\}$ , making the fourth option correct

7 0

2 years ago