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3 years ago

12

Which of the following represents a geometric sequence? PLEASEEEE HELP

2 answers:

monitta3 years ago

7 0

The answer is the second one ☝️

Murljashka [212]3 years ago

3 0

I don’t if i’m right but i think is the second

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When the family arrived in New Delhi from Rome, the youngest son asked the pilot how fast he was flying the plane. The pilot tol

Elden [556K]

Answer:

7 hours

Step-by-step explanation:

It is given that the plane was flying at 847 km/hour. So the speed of the plane was 847 kilometers per hour.

It is also given that the distance between Rome and New Delhi is 5,929 kilometers.

We know that the distance is equal to the product of speed and time, so we have,

$d= s\times t$ , where 'd' represents the distance, 's' represents the speed, and 't' represents the time.

Putting the values we get,

$5929= 847 \times t$

Solving for 't' we get,

$t= \frac{5929}{847}=7$

So the time taken to fly from Rome to New Delhi is 7 hours.

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4 years ago

Who was responsible for inventing the first working four-stroke engine?<br>

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Step-by-step explanation:

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4 years ago

A small pebble has a mass of<br> about<br> 20 L<br> b.<br> 20 ml<br> 20 g<br> 20 kg<br> d.

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3 years ago

A ball is dropped from a height of 12 feet and returns to a height that is one-

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Step-by-step explanation:

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3 years ago

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Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

4 years ago

Read 2 more answers