Step-by-step explanation:
You have studied polynomials consisting of constants and/or variables combined by addition or subtraction. The variables may include exponents. The examples so far have been limited to expressions such as 5x4 + 3x3 – 6x2 + 2x containing one variable, but polynomials can also contain multiple variables. An example of a polynomial with two variables is 4x2y – 2xy2 + x – 7.
Many formulas are polynomials with more than one variable, such as the formula for the surface area of a rectangular prism: 2ab + 2bc + 2ac, where a, b, and c are the lengths of the three sides. By substituting in the values of the lengths, you can determine the value of the surface area. By applying the same principles for polynomials with one variable, you can evaluate or combine like terms in polynomials with more than one variable
Answer:
0.86969
Step-by-step explanation:
1 : 26 soln 1 ÷ 26 1 ÷ 685 now ans 0.86969 the
The answer is D.
First, separate the compound equality into 2
1st -3 (x-4) greater than/ equal to 21
2nd -3( x-4) < 30
Second, solve the inequality for x
1st x less than/ equal to -3
2nd x > -6
Third, find the intersection
D. -6 < x (less than/equal) to -3
I hope this helped, and if you don’t mind, giving me brainiest!
I have no clue how to do that if u do no how to do other math can u help me I need alot of help
Answer:
(x, y) = (1/2, -1)
Step-by-step explanation:
Subtracting twice the first equation from the second gives ...
(2/x +1/y) -2(1/x -5/y) = (3) -2(7)
11/y = -11 . . . . simplify
y = -1 . . . . . . . multiply by y/-11
Using the second equation, we can find x:
2/x +1/-1 = 3
2/x = 4 . . . . . . . add 1
x = 1/2 . . . . . . . multiply by x/4
The solution is (x, y) = (1/2, -1).
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<em>Additional comment</em>
If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.
A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).
