Ask question

Ask question

All categories

3 years ago

6

Which is further 3.345 meter, 3.35 meter of 3.3 meter

1 answer:

Ket [755]3 years ago

6 0

The longest meter is 3.35

You might be interested in

can someone help me with this? this is algebra 2 question and please show me all work (orderd pair, vertex, graph ect..)

Andrews [41]

To find the maximum height you need to find the vertex:(h,k)
Your equation is in vertex form a(x-h)+k and the vertex is (h,k) where k is the maximum height and the h is the distance it went to reach the maximum height.
k=6 so the kangaroo's maximum height is 6 feet.

To find how long is the kangaroo's jump, take a look at the graph. You will notice that the parabola ends at the distance the kangaroo jumped. You will also see that it is the one of the x-intercepts.

-.03(x-14)^2+6=0
-.03(x-14)^2+6-6=0-6
-.03(x-14)^2=-6
-.03/-.03(x-14)=-6/-.03
(x-14)^2=200
[(x-14)^2]^.5=200^.5
x-14=(200)^.5
x-14+14=(200)^.5+14
x≈28.14 feet
The kangaroo jumped a distance of 28.14 feet.

You will notice that the square root of a number gives you two solutions a positive and a negative one. The other solution is -.14, which we know distance is not negative so we do not use that solution. Also, I used the ^.5 instead of using the square root. It is the same.

8 0

4 years ago

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

An airplane has a wingspan of 20 feet and a length of 15 feet. If the model of this airplane has a wingspan of 8 inches, what is

Katen [24]

Answer:

The length of the model is 6 inches.

Step-by-step explanation:

The wingspan of the airplane and length of the airplane are a ratio of 4:3

The model is also supposed to be 4:3.

4:3 x 2 = 8:6

Or, 20x12=240

240/8=30

15x12=180

180/30=6.

8 0

3 years ago

Read 2 more answers

How. Many 5s is in 200989

olga_2 [115]

Answer:

40197 these many 5s make up 200985 so 4 are left to make 200989

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Which integer represents a golf score of 4 over par?

fgiga [73]

Omg ur learning about integers, I'm learning about integers too, what grade are you in?

7 0

3 years ago