Answer:
a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle
Step-by-step explanation:
Let l = the original length of the original rectangle
Let w = the original width of the original rectangle
From the description of the problem, we can construct the following two equations
l=2*w (Equation #1)
(l+4)*(w+4)=l*w+88 (Equation #2)
Substitute equation #1 into equation #2
(2w+4)*(w+4)=(2w*w)+88
2w^2+4w+8w+16=2w^2+88
collect like terms on the same side of the equation
2w^2+2w^2 +12w+16-88=0
4w^2+12w-72=0
Since 4 is afactor of each term, divide both sides of the equation by 4
w^2+3w-18=0
The quadratic equation can be factored into (w+6)*(w-3)=0
Therefore w=-6 or w=3
w=-6 can be rejected because the length of a rectangle can't be negative so
w=3 and from equation #1 l=2*w=2*3=6
I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.
Answer:
a. 
b. View graph
c. 6.40u
Step-by-step explanation:
knowing that the triangle area is equal to base by heigh between two, then:
The length of the longest altitude of your triangle is:
finally it can be seen that the position of the triangle does not matter, as long as the base and heigh are maintained, the area of the triangle will be the same
Answer:
can you post a clearer picture ?
Step-by-step explanation:
Make one side length of the cube s.
V = lwh, but the length, width, and height are all s.
So V = sss, or s^3.
If it has a side length of 2^4, then 2^4 = 16.
V = s^3, or V = 16 times 16 times 16, which = 4096.
