Answer:
T0 =0, T1=0
Step-by-step explanation:
Please kindly go through the attached files for a detailed explanation.
I arranged them from the first one to the last one.
Answer:
Volume of Gold = 0.4 cubic feet.
Step-by-step explanation:
First we have to make sure the units are same as the units the answer is required in.
Answer should be given in cubic feet.
Density is given in pounds per cubic foot. And Mass is given in pounds. So units are in order. Therefore, we can go ahead with the calculation.
The Density of Gold = [Mass of Gold/Volume of Gold]
By substituting values,
ponds per cubic foot = [
pounds / Volume of gold]
Making Volume of God the subject of the equation,
Volume of Gold = [
pounds / tex]1204[/tex] ponds per cubic foot ]
Volume of Gold = 0.4 cubic feet.
Part 1: Answer:
(x+1)(x+1)(x-6) = x^3 - 4x^2 - 11x - 6
Step-by-step explanation:
To make r a root, include (x-r) as a factor. (-1+1)(-1+1)(-1-6) is zero even though (-1-6) isn't.
(6+1)(6+1)(6-6) is zero.
Part 2 Answer:
Standard form: y = -x^4 + 12
Degree 4
left end goes down, right end goes down.
Step by step: apply the definitions of standard form, polynomial degree, and "end behavior". In other words, read the textbook.
Part 3: Answer: x = 3, x = 8
Step by step:
x^2-11x = -24
x^2-11x+24 = 0
(x-3)(x-8) = 0
x = 3 or x = 8
Part 4a Answer:
quotient 2x^2 + x - 3
remainder 1
Step by step:
2x^2 + x - 3
___________________
x-4 ) 2x^3 - 7x^2 - 7x + 13
2x^3 - 8x^2
__________
0 + x^2 - 7x + 13
x^2 - 4x
____________
0 - 3x + 13
- 3x + 12
______
1
Part 4b answer:
quotient 2x^2 - 6x + 2
remainder -20
Step by step: you have to know exactly what you are doing. Refer to textbook or Wikipedia.
dividend 2x^3 +14x^2 - 58x
divisor x+10
leading coefficient of divisor must be 1
write coefficients of dividend at top
write coefficients of dividend at left
| 2 14 -58 0
-10 | -20 60 -20
___________
| 2 -6 2 -20
Coefficients of quotient are 2 -6 2
Remainder is -20
quotient = 2x^2 - 6x + 2
Answer:
15 cups
Step-by-step explanation:
5x3 is 15
even the denominator by multiplying.
do what you multiplied on the numerator.
Answer:
Total area = 237.09 cm²
Step-by-step explanation:
Given question is incomplete; here is the complete question.
Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)
From the figure attached,
Area of the right triangle I = 
Area of ΔADC = 
= 
= 
= 
= 
= 30 cm²
Area of equilateral triangle II = 
Area of equilateral triangle II = 
= 
= 73.0925
≈ 73.09 cm²
Area of rectangle III = Length × width
= CF × CD
= 7 × 5
= 35 cm²
Area of trapezium EFGH = 
Since, GH = GJ + JK + KH
17 = 
12 = 
144 = (81 - x²) + (225 - x²) + 2
144 - 306 = -2x² + 
-81 = -x² + 
(x² - 81)² = (81 - x²)(225 - x²)
x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴
144x² - 11664 = 0
x² = 81
x = 9 cm
Now area of plot IV = 
= 99 cm²
Total Area of the land = 30 + 73.09 + 35 + 99
= 237.09 cm²