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3 years ago

Arthur competed in three different races. His time in the first race was 14.48 seconds, his time in the

Mathematics

2 answers:

jekas [21]3 years ago

8 0

D. 94.72 hope that helps

Step2247 [10]3 years ago

3 0

83.52

Step-by-step explanation:

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Solve the following differential equation: (1− 5 y +x) dy/dx +y= 5/x −1 .<br><br> C=

Ganezh [65]

Answer:

$y(1+x)+x-5\ln xy=C$

Step-by-step explanation:

Consider the given differential equation is

$(1-\frac{5}{y}+x)\frac{dy}{dx}+y=\frac{5}{x}-1$

$(1-\frac{5}{y}+x)\frac{dy}{dx}=\frac{5}{x}-1-y$

$(1-\frac{5}{y}+x)dy=(\frac{5}{x}-1-y)dx$

Taking all variables on right sides.

$(1-\frac{5}{y}+x)dy-(\frac{5}{x}-1-y)dx=0$

$(-\frac{5}{x}+1+y)dx+(1-\frac{5}{y}+x)dy=0$

Let as assume,

$M=-\frac{5}{x}+1+y$ and $N=1-\frac{5}{y}+x$

Find partial derivatives $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y}=1$ and $\frac{\partial N}{\partial x}=1$

Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ , therefore the given differential equation is exact.

The solution of the exact differential equation is

$\int Mdx+\int N(\text{without x)}dy=C$

$\int (-\frac{5}{x}+1+y)dx+\int (1-\frac{5}{y})dy=C$

$yx-5\ln x+x+y-5\ln y=C$

$y+x+xy-5\ln x-5\ln y=C$

$y(1+x)+x-5(\ln x+\ln y)=C$

$y(1+x)+x-5\ln xy=C$

7 0

3 years ago

Question 13 plz show ALL STEPS so I can learn thnx

evablogger [386]

9514 1404 393

Answer:

a) (x³ -x² +x +2) +2/(x+1)

b) (x² +2x -5) +6/(x+3)

Step-by-step explanation:

Polynomial long division is virtually identical to numerical long division, except that the quotient term does not require any guessing. It is simply the ratio of the leading terms of the dividend and divisor. As with numerical long division, the product of the quotient term and the divisor is subtracted from the dividend to form the new dividend for the next step.

The process stops when the dividend is of lower degree than the divisor.

In part (a), you need to make sure the dividend expression has all of the powers of x present. This means terms 0x³ and 0x² must be added as placeholders in the given dividend. They will become important as the work progresses.