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3 years ago

5

3% scored a ball is dropped and bounce to a height of 5 feet. The ball rebounds to 7% of the previous height after each bounce.

The function could be used to be model to admit term in the sequence of the heights of the ball after
drop is

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Does anyone know the answer yet?

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Free_Kalibri [48]

Answer:

<em>Caca</em>

Step-by-step explanation:

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3 years ago

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mash [69]

Answer:

$A)\:x$

$5(x+5)$

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3 years ago

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s344n2d4d5 [400]

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3 years ago

1. A jeepney driver claims that his average monthly income is Php 3000.00 with a standard deviation of Php 300.00. A sample of 3

nydimaria [60]

Answer:

Kindly check explanation

Step-by-step explanation:

The hypothesis :

H0 : μ = 3000

H0 : μ ≠ 3000

The test statistic :

(xbar - μ) ÷ (s/√(n))

xbar = 3500

μ = 3000

σ = 300

n = 30

(3500 - 3000) ÷ (350/√(30))

Test statistic = 7.824

Df = 30 - 1 = 29

Tcritical at 0.01 = 2.462

Test statistic > critical value ; we reject H0 ; and concluded that there is significant evidence that

μ ≠ 3000

7 0

3 years ago

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

8 0

1 year ago