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2 years ago

Explain how to find the original and the new dimensions of an object when the scale changes.

2 answers:

7 0

<span>To find the original dimensions, write a proportion with the scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve. To find new dimensions, write a proportion with the new scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve.</span>

ollegr [7]2 years ago

5 0

Answer:

To find the original dimensions, write a proportion with the scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve. To find new dimensions, write a proportion with the new scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve.

Step-by-step explanation:

This is the sample response on Edg

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Why was the far eastern diplomat at the north pole

n200080 [17]

The reason why the far eastern diplomat was at the north pole was : he got disoriented To travel in the past, people in the past rely on stars and compass. But due to electromagnetic phenomenon, the compass tend to not functioning properly and make it really difficult to actually find out the correct path

6 0

2 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

A running coach wants to know if participating in weekly running clubs significantly improves the time to run a mile. The runnin

patriot [66]

Answer:

Option B is correct.

Use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.

Step-by-step Explanation:

The clear, complete table For this question is presented in the attached image to this solution.

It should be noted that For this question, the running coach wants to test if participating in weekly running clubs significantly improves the time to run a mile.

In the data setup, the mean time to run a mile in January for those that participate in weekly running clubs and those that do not was provided.

The mean time to run a mile in June too is provided for those that participate in weekly running clubs and those that do not.

Then the difference in the mean time to run a mile in January and June for the two classes (those that participate in weekly running clubs and those that do not) is also provided.

Since, the aim of the running coach is to test if participating in weekly running clubs significantly improves the time to run a mile, so, it is logical that it is the improvements in running times for the two groups that should be compared.

Hence, we should use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.

Hope this Helps!!!

7 0

2 years ago

Diana drove at an average speed of 50 miles an hour for 1.5 hours.she then drove at an average speed of 42 miles an hour for 2.2

weqwewe [10]

Diana drove 169.5 miles in total.

Step-by-step explanation:

Average speed first = 50 miles

Driving time = 1.5 hours

Distance = Speed*Time

$Distance=50*1.5\\Distance=75\ miles$

Average speed afterwards = 42 miles

Driving time = 2.25 hours

Distance=Speed*Time

$Distance=42*2.25\\Distance=94.5\ miles$

Total distance = 75+94.5

Total distance = 169.5 miles

Diana drove 169.5 miles in total.

Keywords: distance, speed

Learn more about speed at:

#LearnwithBrainly

4 0

2 years ago

Ishmael bought a brand new bag of green, yellow, and red marbles. There are 40 marbles in the bag. 1/4 of the marbles are yellow

VikaD [51]

Answer:

there are 10 red marbles, 10 yellow marbles and 10 green marbles.

Step-by-step explanation:

think 1/4. the 4 is the same as 40 so if there is 1 that would be 10. so the rest of the colors are all 10 marbles:)

3 0

2 years ago