Ayo can someone help me with this real quick

Marco needs $57 to buy new basketball shoes. If Marco earns $3 per day working and already has $12 saved, which equation shows h

Annette [7]

It’s D

Let me know if you need an explanation!

8 0

3 years ago

Using the law of cosines, write an algebraic proof to show that the angles in an equilateral triangle must equal 60°. Use "∧" to

bekas [8.4K]

The algebraic proof shows that the angles in an equilateral triangle must equal 60° each

<h3>Laws of cosines </h3>

From the question, we are to use the law of cosines to write an algebraic proof that shows that the angles in an equilateral triangle must equal 60°.

Given any triangle ABC, the measures of angles A, B, and C by the law of cosines are

cos A = (b^2 + c^2 - a^2)/2bc

cos B= (a^2 + c^2 - b^2)/2ac

cos C = (a^2 + b^2 - c^2)/2ab

Now, given that the triangle is equilateral, with each of the side lengths equal to s

That is, a = b = c = s

Then, we can write that

cos A = (s^2 + s^2 - s^2)/(2s×s)

cos A = (s^2 )/(2s^2)

cos A = 1/2

cos A = 0.5

∴ A = cos⁻¹(0.5)

A = 60°

Also

cos B = (s^2 + s^2 - s^2)/(2s×s)

cos B = (s^2 )/(2s^2)

cos B = 1/2

cos B = 0.5

∴ B = cos⁻¹(0.5)

B = 60°

and

cos C = (s^2 + s^2 - s^2)/(2s×s)

cos C = (s^2 )/(2s^2)

cos C = 1/2

cos C = 0.5

∴ C = cos⁻¹(0.5)

C = 60°

Thus,

A = 60°, B = 60° and C = 60°

Hence, the algebraic proof above shows that the angles in an equilateral triangle must equal 60° each.

Learn more on The law of cosines here: brainly.com/question/2866347

#SPJ1

3 0

2 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line

Ber [7]

Answer:

The distance is:

$\displaystyle\frac{3\sqrt{142}}{10}$

Step-by-step explanation:

We re-write the equation of the line in the format:

$\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}$

Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.

In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:

$\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right>$

Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:

$x=-2+3t\\y=3+2t\\\displaystyle z=-4+\frac{5}{3}t$

We plug them into the equation of the plane to get the intersection of that line and the plane, since that intersection is the image on the plane of the point (-2, 3, -4) parallel to the given line:

$\displaystyle x+y+z=3\to -2+3t+3+2t-4+\frac{5}{3}t=3$

Then we solve that equation for t, to get:

$\displaystyle \frac{20}{3}t-3=3\to t=\frac{9}{10}$

Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:

$\displaystyle x=-2+3\left(\frac{9}{10}\right)=\frac{7}{10}\\\displaystyle y=3+2\left(\frac{9}{10}\right)=\frac{24}{5} \\\displaystyle z=-4+\frac{5}{3}\left(\frac{9}{10}\right)=-\frac{5}{2}$