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2 years ago

Find the midpoint of points A(-3,7) and B(-6,0) graphically.

Mathematics

2 answers:

Masteriza [31]2 years ago

5 0

Answer:

satingin ko po at B I not sure

Step-by-step explanation:

i hope it help

Bezzdna [24]2 years ago

4 0

Since it says graphically, plot the point (-3/2, 7/2)

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2 years ago

Which of the following functions have a vertical asymptote for values of Ø such that cos Ø = 1?Select all that apply.

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Answer:

$y=\cot \theta$ and $y=\csc \theta$

Step-by-step explanation:

Note that if $\cos \theta=1,$ then $\sin \theta=0.$

Functions $y=\cos \theta,\ \ y=\sin \theta$ do not have vertical asymptotes at all.

Vertical asymptotes have functions $y=\tan \theta,\ \ y=\cot \theta,\ \ y=\sec \theta,\ \ y=\csc \theta.$

Functions $y=\tan \theta$ and $y=\sec \theta$ have the same vertical asymptotes (when $\cos \theta =0$ ).

Functions $y=\cot \theta$ and $y=\csc \theta$ have the same vertical asymptotes (when $\sin \theta =0$ ). See attached diagram

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3 years ago

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2 years ago

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2 years ago

P = √mx_ £x MX t make x the subject

vladimir2022 [97]

So, making x subject of the formula, x = [m - 2pt³ ±√(m² - 4pt²m)]/{2t⁵}

<h3>How to make x subject of the formula?</h3>

Since p = √(mx/t) - t²x

So, p + t²x = √(mx/t)

Squaring both sides, we have

(p + t²x)² = [√(mx/t)]²

p² + 2pt²x + t⁴x² = mx/t

Multiplying through by t,we have

(p² + 2pt²x + t⁴x²)t = mx/t × t

p²t + 2pt³x + t⁵x² = mx

p²t + 2pt³x + t⁵x² - mx = 0

t⁵x² + 2pt³x - mx + p²t = 0

t⁵x² + (2pt³ - m)x + p²t = 0

Using the quadratic formula, we find x.

$x = \frac{-b +/-\sqrt{b^{2} - 4ac} }{2a}$

where

a = t⁵,
b = (2pt³ - m) and
c = p²t

Substituting the values of the variables into the equation, we have

$x = \frac{-(2pt^{3} - m) +/-\sqrt{(2pt^{3} - m)^{2} - 4(t^{5})(p^{2}t) } }{2t^{5} }\\= \frac{-(2pt^{3} - m) +/-\sqrt{4p^{2} t^{6} - 4pt^{2}m + m^{2} - 4p^{2}t^{6} } }{2t^{5}}\\= \frac{-(2pt^{3} - m) +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}\\= \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}$

So, making x subject of the formula, $x = \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}$

Learn more about subject of formula here:

brainly.com/question/25334090

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1 year ago

Find the midpoint of points A(-3,7) and B(-6,0) graphically. ​

Find the midpoint of points A(-3,7) and B(-6,0) graphically.