Renting a small moving truck for a day costs $29.95 plus $0.79 per mile that the truck is driven. Ms. Knowles rented a truck for one day and paid $91.57. The equation 29.95 + 0.79d = 91.57 can be used to model this situation. Choose True or False for each statement. A) The variable d represents the number of miles Ms. Knowles drove. B) The term 0.79d represents the charge in dollars for the distance driven. C) You can isolate the term 0.79d by adding 29.95 to both sides of the equation.

Given:

Renting a small moving truck for a day costs $29.95 plus $0.79 per mile that the truck is driven. Ms. Knowles rented a truck for one day and paid $91.57.

The equation is

To find:

Whether the statement are true or false.

Solution:

We have,

Here,

Fixed cost = 29.95

Variable cost = 0.79d, it depends on the number of miles Ms. Knowles drove.

So, the variable d represents the number of miles Ms. Knowles drove.  Hence, statement A is true.

The term 0.79d represents the charge in dollars for the distance driven. Hence, statement B is true.

You can isolate the term 0.79d by subtracting 29.95 from both sides of the equation.

Hence, statement C is false.

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.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3