Third time im asking since everyone does ajsdfhlaksjd will give brainliest no trolls i need really bad part a: explain why the x-coordinates of the points where the graphs of the equations y = 4^x and y = 2x^−2 intersect are the solutions of the equation 4x = 2x^−2. (4 points) part b: make tables to find the solution to 4^x = 2^x−2. take the integer values of x between −3 and 3. (4 points) part c: how can you solve the equation 4^x = 2^x−2 graphically? (2 points

Part A: We have two lines: y = 2-x and y = 4x+3 . Given twoequations that are both required to be true. The answer is the points where the lines cross... Which is where the two equations are equal. The answer that works for both equations is:

2-x = 4x+3

because where that is true is where the two lines will cross and that is the common point that works for both equations.

Part B:

To find the answer, rearrange the equation to the form x = n

2-x = 4x+3

2 -x + x = 4x + x +3

2 = 5x + 3

2-3 = 5x +3-3

5x = -1

x = -1/5

The only point that works for both equations is where x = -1/5

So find y: y = 2-x = 2 - (-1/5) = 2 + 1/5 = 10/5 + 1/5 = 11/5

Now make sure this is the

y = 4x + 3 = 4(-1/5) + 3 = -4/5 + 15/5 = 11/5

The answer is: (-1/5, 11/5)

bru

step-by-step explanation:

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.