2 years ago

!!!!! PLEASE HELP I'LL FOLLOW WHOEVER GETS IT RIGHT!!!!

Mathematics

1 answer:

g100num [7]2 years ago

4 0

Answer:

Step-by-step explanation:

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2 years ago

Write the equation of a line that is parallel to y = 7x +13 that passes through the point (28, 8).

navik [9.2K]

Answer:

y=7x-188

Step-by-step explanation:

We know that this line must have a slope of 7, because this line is parallel to y=7x+13

We must use the point slope formula: y- $y_{1}$ = m(x- $x_{1}$ )

Next, we will plug in the point (28,8) for $y_{1}$ and $x_{1}$ , and 7 in for m

y-8=7(x-28)

Solve the equation, first by distributing the 7

y-8=7x-196

Add the 8 over

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3 years ago

find the value of the trigonometric function sin (t) if sec t = -4/3 and the terminal side of angle t lies in quadrant II

kodGreya [7K]

Answer:

$sin(t) =\frac{\sqrt{7}}{4}$

Step-by-step explanation:

By definition we know that

$sec(t) = \frac{1}{cos(t)}$

and

$cos ^ 2(t) = 1-sin ^ 2(t)$

As $sec(t) = -\frac{4}{3}$

Then

$sec(t) = -\frac{4}{3}\\\\\frac{1}{cos(t)} =-\frac{4}{3}\\\\cos(t) = -\frac{3}{4}$

Now square both sides of the equation:

$cos^2(t) = (-\frac{3}{4})^2$

$cos^2(t) = \frac{9}{16}\\\\$

$1-sin^2(t) =\frac{9}{16}\\\\sin^2(t) =1-\frac{9}{16}\\\\sin^2(t) =\frac{7}{16}\\\\sin(t) =\±\sqrt{\frac{7}{16}}$

In the second quadrant sin (t) is positive. Then we take the positive root

$sin(t) =\sqrt{\frac{7}{16}}$

$sin(t) =\frac{\sqrt{7}}{4}$

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2 years ago

!!!!! PLEASE HELP I'LL FOLLOW WHOEVER GETS IT RIGHT!!!!​

!!!!! PLEASE HELP I'LL FOLLOW WHOEVER GETS IT RIGHT!!!!