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:
48 cups
Step-by-step explanation:
4 gallons = 64 cups
1 gallon = 64 ÷ 4
1 gallon = 16 cups
3 gallons = 16 * 3 = 48 cups
Answer:
D. (1, -2)
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define systems</u>
5x - 2y = 9
3x + 4y = -5
<u>Step 2: Rewrite systems</u>
10x - 4y = 18
3x + 4y = -5
<u>Step 3: Solve for </u><em><u>x</u></em>
- Add to equations together: 13x = 13
- Divide 13 on both sides: x = 1
<u>Step 4: Solve for </u><em><u>y</u></em>
- Define: 3x + 4y = -5
- Substitute in <em>x</em>: 3(1) + 4y = -5
- Multiply: 3 + 4y = -5
- Isolate <em>y </em>term: 4y = -8
- Isolate <em>y</em>: y = -2
And we have our final answer!
1 23/100 ...............................................
Max occurs at x=50 where F'(x) = 0.
We need to find value for F(50) using fundamental theorem of calculus:
Where A is the estimated Area under the graph.
If we assume the shaded region is close to a right triangle then the Area is (1/2)*30*20 = 300
Therefore
