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3 years ago

I need help on this question

Mathematics

1 answer:

denis23 [38]3 years ago

6 0

Answer:

Step-by-step explanation:

I don't understand the question very well but actually it's a BC line and the other choice it's not correct.

I hope that it's a correct answer.

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Bus A and Bus B leave the bus depot at 9 am. Bus A takes 30 minutes to complete its route once and bus B takes 40 minutes to com

lutik1710 [3]

Answer:

Bus A: 10:00am Bus B:10:20

Step-by-step explanation

I think that this is correct

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3 years ago

I WILL GIVE BRAINLIEST!!! PZ HELP ASAP!!!

CaHeK987 [17]

Answer:

Root of 27 will be 9×3. Which will be 3 root 3

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3 years ago

Read 2 more answers

The square below represents one whole.

sergij07 [2.7K]

Answer:

Decimal: 0.23

Fraction: 23/100

Percent: 23%

Step-by-step explanation:

In this case, one whole is 100. This is because there are 100 little squares in the one big square.

The amount of shaded squares is 23.

This can be represented as the fraction 23/100. 23/100 is the same as 23 being divided by 100. 23 divided by 100 is 0.23 (our decimal). To find a percent, move the decimal point to the right 2 times. Our percent is 23%.

6 0

3 years ago

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5<img src="https://tex.z-dn.net/?f=t%5E%7

DanielleElmas [232]

Answer:

Part A)

Part B)

About 2.9362 seconds.

Step-by-step explanation:

The equation $\displaystyle h(t)=-5t^2+14t+2$ models the height h in meters of the ball t seconds after its launch.

Part A)

To determine whether or not the ball reaches a height of 14 meters, we can find the vertex of our function.

Remember that the vertex marks the maximum value of the quadratic (since our quadratic curves down).

If our vertex is greater than 14, then, at some time t, the ball will definitely reach a height of 14 meters.

However, if our vertex is less than 14, then the ball doesn’t reach a height of 14 meters since it can’t go higher than the vertex.

So, let’s find our vertex. The formula for vertex is given by:

$\displaystyle (-\frac{b}{2a},h(-\frac{b}{2a}))$

Our quadratic is:

$\displaystyle h(t)=-5t^2+14t+2$

Hence: a=-5, b=14, and c=2.

Therefore, the x-coordinate of our vertex is:

$\displaystyle x=-\frac{14}{2(-5)}=\frac{14}{10}=\frac{7}{5}$

To find the y-coordinate and the maximum height, we will substitute this value back in for x and evaluate. Hence:

$\displaystyle h(\frac{7}{5})=-5(\frac{7}{5})^2+14(\frac{7}{5})+2$

Evaluate:

$\displaystyle \begin{aligned} h(\frac{7}{2})&=-5(\frac{49}{25})+\frac{98}{5}+2 \\ &=\frac{-245}{25}+\frac{98}{5}+2\\ &=\frac{-245}{25}+\frac{490}{25}+\frac{50}{25}\\&=\frac{-245+490+50}{25}\\&=\frac{295}{25}=\frac{59}{5}=11.8\end{aligned}$

So, our maximum value is 11.8 meters.

Therefore, the ball doesn’t reach a height of 14 meters.

Part B)

To find out how long the ball is in the air, we can simply solve for our t when h=0.

When the ball stops being in the air, this will be the point at which it is at the ground. So, h=0. Therefore:

$0=-5t^2+14t+2$

A quick check of factors will reveal that is it not factorable. Hence, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Again, a=-5, b=14, and c=2. Substitute appropriately:

$\displaystyle x=\frac{-(14)\pm\sqrt{(14)^2-4(-5)(2)}}{2(-5)}$

Evaluate:

$\displaystyle x=\frac{-14\pm\sqrt{236}}{-10}$

We can factor the square root:

$\sqrt{236}=\sqrt{4}\cdot\sqrt{59}=2\sqrt{59}$

Hence:

$\displaystyle x=\frac{-14\pm2\sqrt{59}}{-10}$

Divide everything by -2:

$\displaystyle x=\frac{7\pm\sqrt{59}}{5}$

Hence, our two solutions are:

$\displaystyle x=\frac{7+\sqrt{59}}{5}\approx2.9362\text{ or } x=\frac{7-\sqrt{59}}{5}\approx-0.1362$

Since our variable indicates time, we can reject the negative solution since time cannot be negative.

Hence, our zero is approximately 2.9362.

Therefore, the ball is in the air for approximately 2.9362 seconds.

5 0

3 years ago

Read 2 more answers

IS this corretct ? THANKYOU! :D WILL REWARD BRAINLIEST ANSWER!

AlladinOne [14]

No, the answers are incorrect.
When reflected over the y-axis, if the x is positive, then it becomes negative, but if the x is negative, then it becomes positive. Also, the y will not change.
The answer should be
X prime: (0,-6)
Y prime: (10,4)
Z prime: (-3,-1)

5 0

3 years ago