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insens350 [35]
3 years ago
12

Is 5/12 greater than or less than 3/6?

Mathematics
1 answer:
Advocard [28]3 years ago
8 0
5/12 is less than 3/6.
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One question simple please help
igor_vitrenko [27]

Answer:

y = \frac{1}{3}x - 9

Step-by-step explanation:

Standard equation of a line is y = mx + b, where m is the slope.

Given line y = - 3x + 78, slope, m₁ = -3

<em><u>To find the line perpendicular to the given line.</u></em>

The lines are perpendicular to each other if the product of their slopes = - 1

That is,

         m_1 \times m_2 = -1

So the slope of new line is

                                                   -3 \times m_2 = -1\\\\m_2 = \frac{-1}{-3} = \frac{1}{3}

Therefore , equation \ of \ new \ line \ \\\\(y - y_1) = m_2(x - x_1) , where \ (x_1 , y_1) = (9, -6 )\\\\(y - (-6))= \frac{1}{3}(x -9)\\\\y + 6 = \frac{1}{3} x - 3\\\\y = \frac{1}{3}x -3-6\\\\y = \frac{1}{3}x -9\\\\

7 0
3 years ago
Can someone check my answers? I'm not sure I have them all right...
Semmy [17]
<span>What is the y-intercept of the plane whose equation is 4x + 5y + z = 20?

(0, 4, 0) 


Which equation has intercepts at X(1, 0, 0), Y(0, 1, 0), and Z(0, 0, 2)?

2x + 2y + z = 2


Which equation is equivalent to x + 3y + z = 3?


4x + 12y + 4z = 12


Which of the following points lies in the plane 3x + 2y + 4z = 12?

(4, 3,2)


Hope these answer the questions. Have a nice day.</span>
3 0
3 years ago
What would be the value of $100 after 10 years if you earn 11 percent interest per year
uysha [10]
10 times 11% is 11 dollars, so 11 times 10 equals 110. Then 110 plus 100 equals 210. So you'd have $210 after ten years with 11 percent interest. 
5 0
3 years ago
What property of equality is X-12=7
Andreyy89
The property of equality equal Nineteen
8 0
3 years ago
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
3 years ago
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