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2 years ago

The scale factor of S to T is 4:1

Mathematics

1 answer:

solong [7]2 years ago

6 0

Answer:

Scale Factor - Finding Sides. L1S1. 3) Scale factor of S to T is 4 : 1. ; x = y = 4) Scale factor of G to H is 1 : 7. ; x = y = 20 yd. 10 yd. S. T.

Step-by-step explanation:

A scale factor in math is the ratio between corresponding measurements of an object and a representation of that object. If the scale factor is a whole number, the copy will be larger. If the scale factor is a fraction, the copy will be smaller.

Hope this helps:) Have a great day!

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A deli sells 2 hot dogs for $2.50. What is the constant of proportionality of dollars per hot dog?

attashe74 [19]

Answer:

$1.25 per hot dog

Step-by-step explanation:

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3 years ago

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Alex has $20 and Ellen has $12 Alex saving three dollars per day and Ellen is saving five dollars per day after how many days wi

FinnZ [79.3K]

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3 years ago

: A personal computer manufacturer buys 36% of its chips from Japan and the rest from the United St...

meriva

Answer: 0.9862

Step-by-step explanation:

Given : The probability that the chips belongs to Japan: P(J)= 0.36

The probability that the chips belongs to United States : P(U)= 1-0.36=0.64

The proportion of Japanese chips are defective : P(D|J)=0.017

The proportion of American chips are defective : P(D|U)=0.012

Using law of total probability , we have

$P(D)=P(D|J)\times P(J)+P(D|U)\times P(U)\\\\\Rightarrow\ P(D)=0.017\times0.36+0.012\times0.64=0.0138$

Thus , the probability that chip is defective = 0.0138

Then , the probability that a chip is defect-free= $1-0.0138=0.9862$

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3 years ago

How would you graph (–4,9)?

Irina-Kira [14]

The answer is B. This is right cause the negatives are to the left and positives up.

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3 years ago

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Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

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3 years ago