TUZ. U410) The table of values represents a linear function g(x), where x is the number of days that have passed and g(x) is the balance in the bank account
x g(x)
$325
5
$400
10 $475
Part A: Find and interpret the slope of the function (3 points)
Part B: Write the equation of the line in point-slope, slope-intercept, and standard forms (3 points)
Part C: Write the equation of the line using function notation (2 points)
Part D: What is the balance in the bank account after 12 days? (2 points)
Question Answered

Part A: 15

Part B:

Point slope form:

Slope intercept form:

Standard Form:

Part C:

Part D: 505

Step-by-step explanation:

We are given the following

x       g(x)

5      $400

10    $475

We can see that these are two points on the line.

i.e. (5, 400) and (10, 475)

OR

Part A: Slope of the function:

So, slope of the line = 15

It has a positive slope and the function will be increasing with the increasing value of 'x'.

Part B:

Point slope form of a line is given as:

Putting :

Slope intercept form:

where c is the intercept.

Putting (x,y) as (5, 400) to find c:

So, the equation in slope intercept form:

Standard form of line:

Rewriting the slope intercept form:

Part C:

Using the function notation, putting y = g(x) in slope intercept form:

Part D:

Balance after 12 days = ?

i.e. g(x) = ? at x = 12

Putting x = 12 in

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.

{3 + i, 3 - i, 3, -1}

step-by-step explanation: