Answer: In your problem you will have to combine liked terms, and you can't combine non-like terms. After you combine your liked terms you will get your answer.
Here is my theory. I did not use the Pythagorean Theorem. I used basic cross multiplication.
z = 25
t = 24
To check which ordered pair (point) is in the solution set of the system of given linear inequalities y>x, y<x+1; we just need to plug given points into both inequalities and check if that point satisfies both inequalities or not. If any point satisfies both inequalities then that point will be in solution.
I will show you calculation for (5,-2)
plug into y>x
-2>5
which is clearly false.
plug into y<x+1
-2<5+1
or -2<6
which is also false.
hence (5,-2) is not in the solution.
Same way if you test all the given points then you will find that none of the given points are satisfying both inequalities.
Hence answer will be "No Solution from given choices".
Answer: C
Step by step: .........
Answer:
1256000
Step-by-step explanation:
My example:
Find the radius of the sphere by substituting 4.5? ft^3 for V in the formula in Step 1 to get: V=4.5? cubic feet.= (4/3)?(r^3)
Multiply each side of the equation by 3 and the equation becomes: 13.5 ? cubic feet =4?(r^3)
Divide both sides of the equation by 4? in Step 4 to solve for the radius of the sphere. To get: (13.5? cubic feet)/(4?) =(4? )(r^3)/ (4?), which then becomes: 3.38 cubic feet= (r^3)
Use the calculator to find the cubic root of 3.38 and subsequently the value of the radius “r” in feet. Find the function key designated for cubic roots, press this key and then enter the value 3.38. You find that the radius is 1.50 ft. You can also use an online calculator for this calculation (see the Resources).
Substitute 1.50 ft. in the formula for SA= 4?(r^2) found in Step 1. To find: SA = 4?(1.50^2) = 4?(1.50X1.50) is equal to 9? square ft.
Substituting the value for pi= ?= 3.14 in the answer 9? square ft., you find that the surface area is 28.26 square ft. To solve these types of problems, you need to know the formulas for both surface area and volume.
