Here's the work:
Density = mass/volume = 129g/(5*3*2) cm^3 = 4.3 g/cm^3.
Answer:
The function is 
domain is 0 to 0.2 hour.
range is 0 to 2 miles.
Step-by-step explanation:
Given that,
Average velocity = 10 miles/hour
Time = 12 min = 0.2 hour
We need to write a function that models the student's distance from home
Using formula of distance

Where, v = velocity
t = time
d = distance
Put the value into the formula

This is a function.
We know that,
Domain :
Domain shows the time.
So, domain = 0 to 0.2 hour
Range :
Range shows the distance.
The range at t =0,
Put the value of t in the function

The range at t =0.2 hour


So. range = 0 to 2 miles
Hence, The function is 
domain is 0 to 0.2 hour
range is 0 to 2 miles
Answer:
perimeter = 28.68 cm
area = 15.48 cm^2
Step-by-step explanation:
1. complete the angles in the triangle. sum of all the angles in a triangle is 180 degrees.
therefore 180 - (60 +45) = 75
angle at B is 75 degrees
2. find the sides of the triangle using the sine formula for triangle
sin A/a = sin B/b = sin C/c
we have the angle at C and the side opposite C is also give, we can use that with any other
3. sin A/a = sin C/c
sin 60/a = sin 45/8
make a subject
a = 9.78 cm
4. sin A/a = sin B/b
sin 60/9.78 = sin 75/b
make b subject
b = 10.90 cm
with the three sides, we know that perimeter is the length around an object
adding all the lengths together will give the perimeter
perimeter = 10.90 + 8 + 9.78
= 28.68 cm
5. to find the area we need to find the high of the triangle since the expression for the area of a triangle is 
6. bisecting the side BC will give have as 4.89 cm
7. using Pythagoras theorem we can find the height of the traingle
c^2 = a^2 + b^2
8^2=(4.89)^2 + b^2
64 - 23.9121 = b^2
b= 
b =6.33 cm
insert this into the formula above will give the value for area
which is 15.48 cm^2
area = 1/2 (4.89)(6.33)
Answer:
L₄₂ = 83.4
Step-by-step explanation:
Put 42 where n is and do the arithmetic.
... L₄₂ = 75 + 0.2·42
... = 75 + 8.4
... = 83.4
Answer:

Step-by-step explanation:
Given



Required
Determine the coordinates of B'
Since A' is the image of A; we need to first calculate the scale factor using

Make Scale Factor the subject

Substitute values for A' and A

Factorize the numerator



The coordinates of B' is then calculated as follows:

Substitute values for B and Scale Factor

