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3 years ago

Calculate without distribution 13.45 × .20 =

Mathematics

2 answers:

amid [387]3 years ago

8 0

Answer:

The answer in this proble is 269

Varvara68 [4.7K]3 years ago

8 0

Answer:

=2.69

Step-by-step

Line up the numbers and multiply up and down

1 3 4 5

x 2

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A function y(t) satisfies the differential equation dy dt = y 4 − 6y 3 + 5y 2 . (a) What are the constant solutions of the equat

Alexxandr [17]

Answer:

Hence increasing in (-\infty,0) U (1,5)

c) Decreasing in (0,1)

Step-by-step explanation:

Given that y(t) satisfies the differential equation

$\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)$

Separate the variables to have

$\frac{dy}{y^2(y-1)(y-5)} =dt$

Left side we can resolve into partial fractions

Let $\frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}$

Taking LCD we get

$1= Ay(y-1)(Y-5) +B(y-1)(y-5)+Cy^2 (y-5)+Dy^2 (y-1)\\Put y =1\\1 = -4C\\Put y =5\\ 1 = 25(4)D\\Put y =0\\1=5B\\$

By equating coeff of y^3 we have

A+C+D=0

$C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}$

Hence left side =

$\frac{6}{25y} +\frac{1}{5y^2}+\frac{-1}{4(y-1)}+ \frac{1}{100(y-5)}=dt\\\frac{6}{25}ln y -\frac{1}{5y}-\frac{1}{4}ln|(y-1)| +\frac{1}{100}ln|y-5| = t+C$

b) y is increasing whenever dy/dt>0

dy/dt =0 at points y =0, 1 and 5

dy/dt >0 in (-\infty,0) U (1,5)

Hence increasing in (-\infty,0) U (1,5)

c) Decreasing in (0,1)

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number of cups you have ÷ number of cups per serving = number of servings

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This is your answer, hope this helps :D !