- Given - <u>a </u><u>rectangle </u><u>with </u><u>length</u><u> </u><u>2</u><u>5</u><u> </u><u>feet </u><u>and </u><u>perimeter </u><u>8</u><u>0</u><u> </u><u>feet</u>
- To calculate - <u>width </u><u>of </u><u>the </u><u>rectangle</u>
We know that ,
where <u>b </u><u>=</u><u> </u><u>width </u><u>/</u><u> </u><u>breadth</u> of rectangle
<u>substituting</u><u> </u><u>the </u><u>values </u><u>in </u><u>the </u><u>formula </u><u>stated </u><u>above </u><u>,</u>
hope helpful ~
The way it's written it's
Simplify the following:
7 X^3 + 4 X^2 + 2 X^2 + 3 X + X + 2 + 5
Grouping like terms, 7 X^3 + 4 X^2 + 2 X^2 + 3 X + X + 2 + 5 = 7 X^3 + (4 X^2 + 2 X^2) + (3 X + X) + (2 + 5):7 X^3 + (4 X^2 + 2 X^2) + (3 X + X) + (2 + 5)
4 X^2 + 2 X^2 = 6 X^2:
7 X^3 + 6 X^2 + (3 X + X) + (2 + 5)
3 X + X = 4 X:
7 X^3 + 6 X^2 + 4 X + (2 + 5)
2 + 5 = 7:Answer: 7 X^3 + 6 X^2 + 4 X + 7
Wouldn't the ratio from length to width be 8?
Answer:x^6-16^2/ -64
Step-by-step explanation:
A plane cuts horizontally across a rectangular pyramid will always be a rectangle. (Answer B)