Which of the following is the graph of the solution set of t – 4 ≥ 4t + 8 or 3t >14 – 4t?

Mathematics

1 answer:

Stells [14]3 years ago

4 0

Answer:

Option (C)

Step-by-step explanation:

There are two inequalities,

t - 4 ≥ 4t + 8 ------(1)

3t > 14 - 4t ------(2)

By simplifying inequality (1),

t - 4 ≥ 4t + 8

t ≥ 4t + 8 + 4

t ≥ 4t + 12

t - 4t ≥ 12

-3t ≥ 12

t ≤ $\frac{12}{(-3)}$

t ≤ -4

By solving inequality (2),

3t > 14 - 4t

3t + 4t > 14

7t > 14

t > $\frac{14}{7}$

t > 2

From the given graphs, Option (C) will be the answer.

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Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

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$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0

3 years ago

6 students want lemonade and 12 students

Ksenya-84 [330]

Answer:

2:1

Step-by-step explanation:

well we show ratio by the sign ' : '

ratio of iced tea to lemonade= 12:6

now we reduce it to minimum by dividing it with same number.

12 is divisible by 6

so,