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stira [4]
2 years ago
10

A7-ft ladder is placed against a wall with its base 3 ft from the wall. How high above the ground is the top of the ladder? Roun

d your answer to the nearest tenth if needed.​
Mathematics
1 answer:
Firdavs [7]2 years ago
4 0

Answer:

6.3 feet

Step-by-step explanation:

use pythagorean theorem

a² + b² = c²

a² + 3² = 7²

a² + 9 = 49

Subtract 9 from both side

a² = 40

Take the square root of both sides

a = 6.324555320336759

Rounded

a = 6.3

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How do you find the derivative of 1/logx?
valentina_108 [34]

The derivative of 1/logx is With the chain rule.

1log(x)=log(x)−1 is ,= -1xlog(x)2 .

The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".

The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

With the chain rule. \\ 1log(x)=log(x)−1  \\ So you follow the chain rule to the electricity after which to the log: \\ ddxlog(x)−1=−1⋅log(x) \\  \\−2⋅ddxlog(x)  \\ =−1log(x)21x  \\ =−1xlog(x)2 .

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5 0
9 months ago
Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.
jarptica [38.1K]
  • y''-y'+y=\sin x

The corresponding homogeneous ODE has characteristic equation r^2-r+1=0 with roots at r=\dfrac{1\pm\sqrt3}2, thus admitting the characteristic solution

y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x

For the particular solution, assume one of the form

y_p=a\sin x+b\cos x

{y_p}'=a\cos x-b\sin x

{y_p}''=-a\sin x-b\cos x

Substituting into the ODE gives

(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x

-b\cos x+a\sin x=\sin x

\implies a=1,b=0

Then the general solution to this ODE is

\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}

  • y''-3y'+2y=e^x\sin x

\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2

\implies y_c=C_1e^x+C_2e^{2x}

Assume a solution of the form

y_p=e^x(a\sin x+b\cos x)

{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)

{y_p}''=2e^x(a\cos x-b\sin x)

Substituting into the ODE gives

2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x

-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x

\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12

so the solution is

\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}

  • y''+y=x\cos(2x)

r^2+1=0\implies r=\pm i

\implies y_c=C_1\cos x+C_2\sin x

Assume a solution of the form

y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)

{y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)

Substituting into the ODE gives

(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)

-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)

\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49

so the solution is

\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}

7 0
3 years ago
Jim wants to put a wallpaper border around a rectangular room. The
puteri [66]

Answer:3

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Select True or False for each statement.
Margarita [4]
The first statement and the second statement are FALSE.

In mathematics, a function<span> is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the </span>function<span> that relates each real number x to its square x</span>2<span>.
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I hope my answer has come to your help. God bless you and have a nice day ahead!
7 0
3 years ago
Hurry please find the values of x and y
MrRissso [65]
Answer:
Y=70
X= 20
You do first the adjacent angles 180-40 so 140 then u have that the triangle is isosceles so the bas angles are equal 180 -140 = 40
40 divided by 2 20 and 20 so x= 20
Then 90 -20 is 70
So 180 - (70+40)
180-110=70 so y= 70
Hope you understand
8 0
2 years ago
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