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3 years ago

B. 17695.137 +2762.8796 -

Mathematics

1 answer:

EleoNora [17]3 years ago

3 0

Answer:

20458.0166

Step-by-step explanation:

175695.137+2742.8796=

20458.0166

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Round 156.5934066 to the nearest tenth

inna [77]

Since this number ends with a .59, this means that we should round 156.59 to 157.00. But because we never use .00 unless we are dealing with money, it's just better to say 157 instead of 157.00.

Hope this helps!

7 0

3 years ago

Read 2 more answers

(2x^3-3x^2-18x-8) / (x-4)

eimsori [14]

<u>2x³ - 3x² - 18x - 8</u> = 2x² + 5x + 2
x - 4

7 0

3 years ago

ms. Omar runs the school tennis club. She has a bin of tennis balls and rackets. For every 5 tennis balls in the bin, there are

anygoal [31]

Answer:

see the explanation

Step-by-step explanation:

Let

x ----> number of tennis balls

y ----> number of tennis rackets

we know that

To find out the ratio of tennis balls to tennis rackets, divide the number of tennis ball by the number of tennis rackets

In this problem we have

$\frac{x}{y}=\frac{5}{3}$

see the model in the attached figure N 1

$y=\frac{3}{5}x$

$y=0.60x$

This is the equation of a linear direct variation

The relationship between the variables x and y is proportional

see the model in the attached figure N 2

4 0

3 years ago

Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second lin

Anastasy [175]

Suppose $y_2(x)=y_1(x)v(x)$ is another solution. Then

$\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}$

Substituting these derivatives into the ODE gives

$x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0$

$x^5v''+5x^4v'=0$

Let $u(x)=v'(x)$ , so that

$\begin{cases}u=v'\\u'=v''\end{cases}$

Then the ODE becomes

$x^5u'+5x^4u=0$

and we can condense the left hand side as a derivative of a product,

$\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C$

$x^5u=C\implies u=Cx^{-5}$

Solve for $v$ :

$v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2$

Solve for $y_2$ :

$\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}$

So another linearly independent solution is $y_2=\dfrac1x$ .

3 0

3 years ago

Question is in the picture

Basile [38]

Answer:

(6, 5)

Step-by-step explanation:

-3 + 9 = 6

15 - 10 = 5

4 0

3 years ago

B. 17695.137 +2762.8796 -​

B. 17695.137 +2762.8796 -