Which set of points includes all of the solutions for y = negative four fifths times x plus two? line going through 0 comma 2 and 5 comma negative 2
(15, 10), (0, 2), and (−10, 10)
(−10, 10), (−5, 6), (0, 2), (5, −2), (10, −6)
(x, y) for all real numbers
(x, negative four fifths times x plus two) for all real numbers

d) for all Real Numbers

Step-by-step explanation:

The given equation is:

We have to find which of the given options list all of the possible solutions for the given equation. i.e. we have to find the option which lists all the points that satisfy the given equation.

Since, there is no restriction on x and y, this gives us a hint that there are an infinite number of values that would  satisfy this equation. So this eliminates the first two options.

Since, the given equation is not an identity, it will not hold true for each and every real value of x. Although, the number of solutions are infinite but it does not include each and every real number. So this eliminates option c as well.

This leaves us with option d.

The first coordinate of the ordered pair is x and the second coordinate is the equation defined for y. i.e. the ordered pair is:

For all real values of x.

For every value of x, the ordered pair obtained will satisfy the given equation as the ordered pair is derived from the original equation. So, this option includes all the solutions for the given equation.

(D) The correct answer is:

for all real numbers.

Step-by-step explanation:

Given information:

The equation

We need to find the possible solutions for the given equations.

As one can see there is no restrictions on x and y, this give us a hint that there are infinite values which will satisfy the given equation.

Since the equation provided is not an identity hence it will not be true for holding each and every real value of x. the infinite solution is there but it does not mean it will hold every possible value of x or each and every real number.

Hence, For the value of x the ordered pair obtained will satisfy the given equation Hence the true solution is:

Hence, the option (d) will be the correct answer.

for all real numbers.

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option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

£56

step-by-step explanation: