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anygoal [31]
3 years ago
13

Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run.

Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0

4.0 seconds you just said the answer

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Simplify 5 (6-3)+2
KATRIN_1 [288]

Answer:

C)

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
True or false? Do the following set of ordered pairs represent a function {(8,1),(0,-1),(9,4),(-6,-7),(8,9)}
Anika [276]
Hi there!

Unfortunately, the set of ordered pairs does NOT represent a function! This is because there are two of the same x values. In a function, there is one input for every output. In this case, there are two of the same inputs. However, there can be two of the same outputs. 
Just to clarify - Not a function

Hope this helps!! :)
If there's anything else that I can help you with, please let me know! 
6 0
3 years ago
Please help me I have a lot of hw please 11=f-16
stepladder [879]

Answer:

f = 27

Step-by-step explanation:

11 = f - 16

Add 16 to both sides.

11 + 16 = f - 16 + 16

27 = f

7 0
3 years ago
Read 2 more answers
The plate is rotated 90° about the origin in the counterclockwise direction. In the image trapezoid, what are the coordinates of
Jet001 [13]

Answer:

The coordinates of the endpoints of the side congruent to side EF is:

E'(-8,-4) and F'(-5,-7).

Step-by-step explanation:

<em>" when point M (h, k) is rotated about the origin O through 90° in anticlockwise direction or we can say counter clockwise. The new position of point </em><em>M (h, k) will become M' (-k, h) "</em>

We are given a trapezoid such that the vertices of trapezoid are:

E(-4,8) , F(-7,5) , G(-4,3) , H(-2,5)

Then the new coordinates after the given transformation is:

E(-4,8) → E'(-8,-4)

F(-7,5) → F'(-5,-7)

G(-4,3) → G'(-3,-4)

H(-2,5) → H'(-5,-2)

Hence the coordinates of the endpoints of the side congruent to side EF is:

E'(-8,-4) and F'(-5,-7).


4 0
3 years ago
Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying
pashok25 [27]

Answer:

a.w(t)=-12e^{2t}

b.y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

(D^2-2D-35)=0

By factorization method we are  finding the solution

D^2-7D+5D-35=0

(D-7)(D+5)=0

Substitute each factor equal to zero

D-7=0  and D+5=0

D=7  and D=-5

Therefore ,

General solution is

y(x)=C_1e^{7t}+C_2e^{-5t}

Let y_1=e^{7t} \;and \;y_2=e^{-5t}

We have to find Wronskian

w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}

Substitute values then we get

w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}

w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}

w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}

a.w(t)=-12e^{2t}

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

y(0)=C_1+C_2

C_1+C_2=-7....(equation I)

y'(t)=7C_1e^{7t}-5C_2e^{-5t}

y'(0)=7C_1-5C_2

7C_1-5C_2=23......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminateC_1

Then we get C_2=-\frac{5}{2}

Substitute the value of C_2 in  I equation then we get

C_1-\frac{5}{2}=-7

C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}

Hence, the general solution is

b.y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}

7 0
3 years ago
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