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lesya692 [45]
3 years ago
6

What's the solution to these questions?

Mathematics
1 answer:
Ann [662]3 years ago
6 0
√(X-2 )+1=5
√(X-2 )=5-1
√(X-2 )=4                        <===     square the tow sides 

X-2  =16
x=16+2
x=18
 ..................................................................
(4x+1)^1/4-4=-1
(4x+1)^1/4=-1+4
(4x+1)^1/4=3
4x+1=(3)^4
4x+1=81
4x=81-1
4x=80
x=80/4
 x=20





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Explain what terms and degrees are and how. How are they used to classify polynomials? Also explain why this polynomial (4x2y +
user100 [1]
When a polynomial has more than one variable, we need to look at each term. Terms are separated by + or - signs. Find the degree of each term by adding the exponents of each variable in it. <span>The degree of the polynomial is found by looking at the term with the highest exponent on its variables.
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Polynomials can be classified in two different ways - by the number of terms and by their degree. A monomial is an expression with a single term. It is a real number, a variable, or the product of real numbers and variables. A polynomial is a monomial or the sum or difference of monomials. A polynomial can be arranged in ascending order, in which the degree of each term is at least as large as the degree of the preceding term, or in descending order, in which the degree of each term is no larger than the degree of the preceding term. 

The polynomial (4x^2y + 5xy) is classified as a 3rd degree binomial, because the monomial 4x^2y has degree equal to 3 and the monomial 5xy has degree equal to 2. The highest degree is 3, therefore the polynomial (4x^2y + 5xy) is classified as a 3rd degree polynomial. Since polynomial <span><span> (4x^2y + 5xy)</span> has two terms, then it is classified as binomial.</span>






6 0
3 years ago
Need Answered ASAP
kicyunya [14]

Answer:

The possible rational roots are: +1, -1 ,+3, -3, +9, -9

Step-by-step explanation:

The Rational Root Theorem tells us that the possible rational roots of the polynomial are given by all possible quotients formed by factors of the constant term of the polynomial (usually listed as last when written in standard form), divided by possible factors of the polynomial's leading coefficient. And also that we need to consider both the positive and negative forms of such quotients.

So we start noticing that since the leading term of this polynomial is x^3, the leading coefficient is "1", and therefore the list of factors for this is: +1, -1

On the other hand, the constant term of the polynomial is "9", and therefore its factors to consider are: +1, -1 ,+3, -3, +9, -9

Then the quotient of possible factors of the constant term, divided by possible factor of the leading coefficient gives us:

+1, -1 ,+3, -3, +9, -9

And therefore, this is the list of possible roots of the polynomial.

6 0
2 years ago
A motorboat travels 275 kilometers in 5 hours going upstream. It travels 405 kilometers going downstream in the same amount of t
VladimirAG [237]

Step-by-step explanation:

275 \times 5 = 1375

The rate of the motorboat on still water is 1375

405 \times 5 = 2025

The current rate of the motorboat is 2025

4 0
3 years ago
Which of the sums below can be expressed as 6(3 + 9)? (3 points) 18 + 54 18 + 9 9 + 15 6 + 54
vitfil [10]

6*3 = 18

6*9 = 54

so 18 +54 is the correct answer

7 0
2 years ago
Read 2 more answers
Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu
Alisiya [41]
Let y=C_1x+C_2x^3=C_1y_1+C_2y_2. Then y_1 and y_2 are two fundamental, linearly independent solution that satisfy

f(x,y_1,{y_1}',{y_1}'')=0
f(x,y_2,{y_2}',{y_2}'')=0

Note that {y_1}'=1, so that x{y_1}'-y_1=0. Adding y'' doesn't change this, since {y_1}''=0.

So if we suppose

f(x,y,y',y'')=y''+xy'-y=0

then substituting y=y_2 would give

6x+x(3x^2)-x^3=6x+2x^3\neq0

To make sure everything cancels out, multiply the second degree term by -\dfrac{x^2}3, so that

f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y

Then if y=y_1+y_2, we get

-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0

as desired. So one possible ODE would be

-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0

(See "Euler-Cauchy equation" for more info)
6 0
3 years ago
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