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3 years ago

6

a 15 foot ladder is leaning against a wall. the bottom of the ladder is 5 feet from the wall. to the nearest foot, how far up th

e wall does the ladder reach!! please help!! and show work

1 answer:

denis23 [38]3 years ago

6 0

The formula for this would be the Pythagorean Theorem. This meaning a²+b²=c²
so a²+15²=25, 15²=225, so a²+225=25, move the 225 over and a²=250, square root of 250 is 15.8 rounded up to 16 feet so, so a=16. but idk that might not be right

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

What is the range of the relation? {(1, 2), (2, 4), (3, 2), (4, 6)} A.{2, 4} B.{1, 2, 3, 4, 6} C.{2, 4, 6} D.{1, 2, 3, 4}

tamaranim1 [39]

$(1)\ \ \ \ range:\ \ \ \{2, 4, 6\}\ \ \ \Rightarrow\ \ \ Ans.\ C\\\\(2)\ \ \ function:\ \ \ \{(1, 2); (2, 2); (3, 2); (4, 2)\}\ \ \ \Rightarrow\ \ \ Ans.\ D\\\\(3)\ \ \ f(4)=2-3\cdot4=2-12=-10\ \ \ \Rightarrow\ \ \ Ans.\ C\\\\(4)\ \ \ direct:\ \ \ y=ax\ \ \ \Rightarrow\ \ \ y= \frac{1}{3} x\ \ \ \Rightarrow\ \ \ Ans.\ B\\\\(5)\ \ \ f(x)=ax\ \ \ and\ \ \ f(4)=12\\.\ \ \ \ \ \Rightarrow\ \ \ 12=a\cdot4\ \ \ \Rightarrow\ \ \ a=3\ \ \ \Rightarrow\ \ \ f(x)=3x\ \ \ \Rightarrow\ \ \ Ans.\ (?)\\\\$

$(6)\ \ \ f(x)=ax^2\ \ \ and\ \ \ f(2)=48\\.\ \ \ \Rightarrow\ \ 48=a\cdot2^2\ \ \Rightarrow\ \ a=48:4=12\ \ \Rightarrow\ \ f(x)=12x^2\ \Rightarrow\ \ Ans.\ C\\\\(7)\ \ \ inversely:\ \ \ f(x)= \frac{a}{x} \ \ \ and\ \ \ f(2)=4\\.\ \ \ \Rightarrow\ \ \ 4= \frac{a}{2} \ \ \ \Rightarrow\ \ \ a=4\cdot2=8\ \ \ \Rightarrow\ \ \ y= \frac{8}{x}\ \ \ \Rightarrow\ \ \ Ans.\ B$

5 0

3 years ago

2y + 12 = -3x<br><br> solve pls i have more I WILL GIVE EXTRA POINTS!!!!!!!!!!<br> 20+

Mrac [35]

2y + 12 = -3x

Step-by-step explanation:

Divide y y by 1 1 . Divide 12 12 by 2 2 . Use the slope-intercept form to find the slope and y-intercept. The slope-intercept form is y=mx+b y = m x + b , where m m is the slope and b b is the y-intercept.

8 0

2 years ago

It costs $10 to get into a carnival. It costs an

LenKa [72]

Answer:

13

Step-by-step explanation:

36-10=$26 spent on rides

26/2=13

4 0

2 years ago

Read 2 more answers

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Stella [2.4K]

I hope this helps you

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2 years ago