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pav-90 [236]
2 years ago
8

Solve for t t/5/12=-3/10

Mathematics
1 answer:
natka813 [3]2 years ago
6 0
T = -18 hope this helped
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Solve 2(b − 3) ≥ 3(3 − b) and describe the graph of the solution. Show all your work.
solong [7]

Answer:

\large\boxed{b\geq3\to b\in[3,\ \infty)}

Step-by-step explanation:

2(b-3)\geq3(3-b)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\(2)(b)+(2)(-3)\geq(3)(3)+(3)(-b)\\\\2b-6\geq9-3b\qquad\text{add 6 to both sides}\\\\2b-6+6\geq9+6-3b\\\\2b\geq15-3b\qquad\text{add}\ 3b\ \text{to both sides}\\\\2b+3b\geq15-3b+3b\\\\5b\geq15\qquad\text{divide both sides by 5}\\\\\dfrac{5b}{5}\geq\dfrac{15}{5}\\\\b\geq3

,\ \geq-\text{line to the right}\\-\text{o}\text{pen circle}\\\leq,\ \geq-\text{closed circle}

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3 years ago
Which angle has a sine of -1/2 and a cosine of -√3/2
Sergeeva-Olga [200]
\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)

\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}

\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}

x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}
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HOW CAN ARITHMETIC AND GEOMETRIC SEQUENCES BECOME LINEAR AND EXPONENTIAL FUNCTIONS?
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A sequence is a set of numbers, called terms, arranged in some particular order. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference. A geometric sequence is a sequence with the ratio between two consecutive terms constant.
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2 years ago
Select the factors of x2 - 14x + 49
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(X-7)(x-7)
Hope this is what you're looking for!
6 0
2 years ago
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2. Simplify the following expression 1 (16 (16-12)​
nikklg [1K]

Answer:

The answer is 64.

16-12=4.

4×16×1=64.

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3 years ago
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