Answer:
The answer is 5.30 your welcome
Answer:
what are the possible inequalities?
Step-by-step explanation:
We write the equation in terms of dy/dx,
<span>y'(x)=sqrt (2y(x)+18)</span>
dy/dx = sqrt(2y + 18)
dy/dx = sqrt(2) ( sqrt(y + 9))
Separating the variables in the equation, we will have:
<span>1/sqrt(y + 9) dy= sqrt(2) dx </span>
Integrating both sides, we will obtain
<span>2sqrt(y+9) = x(sqrt(2)) + c </span>
<span>where c is a constant and can be determined by using the boundary condition given </span>
<span>y(5)=9 : x = 5, y = 9
</span><span>sqrt(9+9) = 5/sqrt(2) + C </span>
<span>C = sqrt(18) - 5/sqrt(2) = sqrt(2) / 2</span>
Substituting to the original equation,
sqrt(y+9) = x/sqrt(2) + sqrt(2) / 2
<span>sqrt(y+9) = (2x + 2) / 2sqrt(2)
</span>
Squaring both sides, we will obtain,
<span>y + 9 = ((2x+2)^2) / 8</span>
y = ((2x+2)^2) / 8 - 9
Answer:
6
Step-by-step explanation:
the mean is 6 so is the median, can I have brainliest plz and thanks if yes
Answer:
Infinite pairs of numbers
1 and -1
8 and -8
Step-by-step explanation:
Let x³ and y³ be any two real numbers. If the sum of their cube roots is zero, then the following must be true:
![\sqrt[3]{x^3}+ \sqrt[3]{y^3}=0\\ \sqrt[3]{x^3}=- \sqrt[3]{y^3}\\x=-y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%2B%20%5Csqrt%5B3%5D%7By%5E3%7D%3D0%5C%5C%20%5Csqrt%5B3%5D%7Bx%5E3%7D%3D-%20%5Csqrt%5B3%5D%7By%5E3%7D%5C%5Cx%3D-y)
Therefore, any pair of numbers with same absolute value but different signs fit the description, which means that there are infinite pairs of possible numbers.
Examples: 1 and -1; 8 and -8; 27 and -27.