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Lelechka [254]
4 years ago
15

Rami has swimming lessons every 3 days and guitar lessons every 8 days. If he has both lessons on the first day if the month, in

his many days will Rami have both lessons on the same day again?
Mathematics
2 answers:
kirill115 [55]4 years ago
8 0

Answer:

3 and 8 do not have common factors so the next time these two lessons would be at the same time is 3*8.

3*8=24

24 days from now he will have both

Step-by-step explanation:


OleMash [197]4 years ago
3 0

Answer:

First you should multiply 8x3, since there are no common factors. Your awnser should be 24 since 8x3=24

Step-by-step explanation:

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Sliva [168]
<span><span>36x2^-30x+25=0, maybe.</span></span>
6 0
3 years ago
The temperature was 65 degrees at daybreak. Then it dropped two degrees per hour until dusk. This decrease in temperature can be
Montano1993 [528]

The temperature 6 hours after day break is 53 degrees

Step-by-step explanation:

The temperature was 65 degrees at daybreak. Then it dropped two degrees per hour until dusk. This decrease in temperature can be modeled by the equation,

y = -2x + 65

according to this equation

y= temperature after day break

x= no of hours ( in this case 6)

So equation becomes

y= -2 (6) +65

y=  65- 12

y=   53 degrees

So the temperature 6 hours after day break is 53 degrees

Keywords: Algebra

Learn more about algebra at:

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#learnwithBrainly

8 0
4 years ago
-. What is 42 increased by .08%?​
Anettt [7]

Answer:

\large\boxed{45.36}

Step-by-step explanation:

In this question, we're trying to find to product after the increase of .08%

.08 is the same as 8%, so we know it's going to increase by 8%

In order to get our answer, we would multiply 42 by 1.08

Solve:

42*1.08=45.36\\\\\text{Or you can do it this way}\\\\42*.08=3.36\\\\42+3.36=45.36\\\\\text{Either method will still get you 45.36}

When you're done solving, you should get 45.36

This means that 45.36 would be your final answer.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
6 0
4 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)

No

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
Find the product of 3y^3(6y^2-9y+6)
Valentin [98]

Answer: Simplified expression- 18y^{5} - 27y^{4} + 18y^{3}

Step-by-step explanation: You get the product by simplifying it.

I tried my best, I hope this it what you were looking for:)

7 0
3 years ago
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