140 pounds * 6 feet = 112 pounds * x feet
x feet = (140 * 6) / 112
x = 7.5 feet
Step-by-step explanation:
Total Surface area of a pyramid is given as
= Lateral surface area + area of base.
Lateral surface area = 1/2×perimeter of base×slant height
given the base is square with side 9 in , and slant height of 12 inches.
therefore, LSA = 1/2(9*4)×12 = 216 in^2
Now, TSA = LSA + Base area = 216+9^2 = 297 in^2
a) It is given that the pyramid is smaller therefore, it will require lesser paint than the original pyramid ( The TSA of the smaller pyramid cannot be calculated as its dimentions are not given).
b) It can not be solved as dimensions are missing in the question.
However, the logic has been explained in the soluion one easily put values to find the solution.
The surface area of a rectangular prism is always:
A=2(xy+xz+yz), where x, y, and z are the three dimensions of the prism.
A=2(50*90+50*3.5+90*3.5)
A=2(4500+175+315)
A=2(4990)
A=9980 mm^2
Option A: z + 1
Option B: 6 + w
Option D: 
Solution:
Let us first define the polynomial.
A polynomial can have constants, variables, exponents and fractional coefficients.
A polynomial cannot have negative exponents, fractional exponents and never divided by a variable.
<u>To find which expressions are polynomial:</u>
Option A: z + 1
By the definition, z + 1 is a polynomial.
It is polynomial.
Option B: 6 + w
By the definition, 6 + w is a polynomial.
It is polynomial.
Option C: ![y^{2}-\sqrt[3]{y}+4](https://tex.z-dn.net/?f=y%5E%7B2%7D-%5Csqrt%5B3%5D%7By%7D%2B4)
![y^{2}-\sqrt[3]{y}+4=y^{2}-{y}^{1/3}+4](https://tex.z-dn.net/?f=y%5E%7B2%7D-%5Csqrt%5B3%5D%7By%7D%2B4%3Dy%5E%7B2%7D-%7By%7D%5E%7B1%2F3%7D%2B4)
Here, y have fractional exponent.
So, it is not a polynomial.
Option D: 
By the definition,
is a polynomial.
It is polynomial.
Hence z + 1, 6 +w and
are polynomials.
Answer:
2, 240
Step-by-step explanation:
In the table, the values should be
Year 1 725
Year 2 579
Year 3 696
Since Year 4 sold 112% of the previous 3 years combined, we add 725+579+696=2000. We can find 112% by writing a proportion.

We solve through cross multiplication of numerator and denominator of the opposite fraction.
