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3 years ago

Solve each proportion 5/t = t/45

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

7 0

$45t\cdot \frac { 5 }{ t } =\frac { t }{ 45 } \cdot 45t\\ \\ 225={ t }^{ 2 }\\ \\ t=\pm \sqrt { 225 } \\ \\ t=\pm 15$

insens350 [35]3 years ago

5 0

T = 15 or -15 because you solve proportions by cross multiplying. if you write it out in the following way:

5 t
__= __
t 45

Then, cross multiplying involves multiplying the numerator( number at the top) of one of the fractions with the denominator (number at the bottom) of the other, and setting the two results equal to each other. You get t * t = 5 * 45 and end up with t^2 = 225 or t = (15 or -15)

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olga_2 [115]

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4 years ago

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DochEvi [55]

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Read 2 more answers

If sin theta = 2/3 and sex theta < 0 , find cos theta and tan theta

FromTheMoon [43]

Answer:

$\displaystyle cos\theta=-\frac{\sqrt{5}}{3}$

$\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

Trigonometric Formulas

To solve this problem, we must recall some basic relations and concepts.

The main trigonometric identity relates the sine to the cosine:

$sin^2\theta+cos^2\theta=1$

The tangent can be found by

$\displaystyle tan\theta=\frac{sin\theta}{cos\theta}$

The cosine and the secant are related by

$\displaystyle cos\theta=\frac{1}{sec\theta}$

They both have the same sign.

The sine is positive in the first and second quadrants, the cosine is positive in the first and fourth quadrants.

The sine is negative in the third and fourth quadrants, the cosine is negative in the second and third quadrants.

We are given

$\displaystyle sin\theta=\frac{2}{3}$

Find the cosine by solving

$sin^2\theta+cos^2\theta=1$

$\displaystyle \left(\frac{2}{3}\right)^2+cos^2\theta=1$

$\displaystyle cos^2\theta=1-\left(\frac{2}{3}\right)^2=1-\frac{4}{9}=\frac{5}{9}$

$\displaystyle cos\theta=\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}$

$\boxed{\displaystyle cos\theta=-\frac{\sqrt{5}}{3}}$

We have placed the negative sign because we know the secant ('sex') is negative and they both have the same sign.

Now compute the tangent

$\displaystyle tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}$

Rationalizing

$\displaystyle tan\theta=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}$

$\boxed{\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}}$

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4 years ago