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3 years ago

1) A group of 32 students and 8 adults are going on a field trip to the recycling center

Mathematics

1 answer:

Marrrta [24]3 years ago

7 0

Answer:

what's the question exactly?

Step-by-step explanation:

it's only half of the question

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A sequence of transformations was applied to an equilateral triangle in a coordinate plane. The transformations used were rotati

Svetlanka [38]

Answer:

A- It must be an equilateral triangle with the same side lengths as the original triangle

Step-by-step explanation:

Well The Equilateral Triangle dont change at all But its position on the coordinate plane changes, So it must be still a Equilateral Triangle.

The other options are talking about how the equilateral triangle changes, But it not true only the positions on the coordinate plane changes.

6 0

3 years ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

Read more about Laplace tranoform

brainly.com/question/14487937

#SPJ4

6 0

2 years ago

Which of the following could not be points on the unit circle?

NemiM [27]

A. √(0.8^2) + (0.6^2) = √1 = 1 => OK

<span>b.(-2/3,√ 5/3) = √(-2/3)^2 + 5/9) = √(4/9 +5/9) = √1 = 1 => OK

c.(√ 3/2, 1/3) = √(3/4 + 1/9) < 1 => it is inside the unit circle

d.(1,1) = √(1 + 1) = √2 > 1 => NO. This point is beyond the limits of the unit circle.</span>

4 0

3 years ago

Read 2 more answers

Can any one help with y = 2x + 4

geniusboy [140]

Answer:

It's in the screenshot hope it helpsss

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Author answered 85 out of 95 questions correctly on his math exam.Miguel answered 64 out of 76 questions correctly on his histor

photoshop1234 [79]

In this case, you need to find the percentage each student got, then find who received a higher score.

First Step: $\frac{85}{95}$ (and) $\frac{64}{76}$
Second step: Divide each fraction → ≈ 0.8947 (and) ≈ 0.8421
Third Step: Multiply each decimal by 100. → 89.47% (and) 84.21%

Fourth Step: As 89.47% is greater, Author got an higher score.

BRAINLIEST PLS!!!

3 0

4 years ago