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harkovskaia [24]
2 years ago
6

Please help me real quick

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
3 0

Answer:

Are you asking for both?

Step-by-step explanation:

I have both answers so which one should i do/help with.

Please explain properly.

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Need help with these table what do I do?
Greeley [361]

Answer:

just ask it again

Step-by-step explanation:

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2 years ago
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A store has 300 televisions on order, and 80% are high definition, how many televisions on order are high difinition?
Brilliant_brown [7]

Answer:240

Step-by-step explanation:

300*80%=240

5 0
3 years ago
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Gerardo has just invested in a bond worth $470 that is expected to increase in value by 6% per year. Write the equation used to
Viefleur [7K]

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $470

r = 6% = 6/100 = 0.06

n = 1 because it was compounded once in a year.

Therefore, the equation used to determine the value of his bond after t years is

A = 470(1 + 0.06/1)^1 × t

A = 470(1.06)^t

3 0
3 years ago
I need help on how to find out this answer asap !!!
amm1812
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6 0
3 years ago
At age 7, Maui competed in the high jump competition and managed a personal best of 3ft. At age 11 she was jumping 5ft. Her coac
Helen [10]

The high-jumper's centre of mass is about two-thirds of the way up his body when he is standing or running in towards the take off point. He needs to increase his launch speed to the highest possible by building up his strength and speed, and then use his energy and gymnastic skill to raise his centre of gravity by H, which is the maximum that the formula U2=2gH will allow. Of course there is a bit more to it in practice! When a high jumper runs in to launch himself upwards he will only be able to transfer a small fraction of his best possible horizontal sprinting speed into his upward launch speed. He has only a small space for his approach run and must turn around in order to take off with his back facing the bar. The pole vaulter is able to do much better. He has a long straight run down the runway and, despite carrying a long pole, the world's best vaulters can achieve speeds of close to 10 metres per second at launch. The elastic fibre glass pole enables them to turn the energy of their horizontal motion 12MU2 into vertical motion much more efficiently than the high jumper. Vaulters launch themselves vertically upwards and perform all the impressive gymnastics necessary to curl themselves in an inverted U-shape over the bar,sending their centre of gravity as far below it as possible.

Pole vaulter

Let's see if we can get a rough estimate of how well we might expect them to do. Suppose they manage to transfer all their horizontal running kinetic energy of 12MU2 into vertical potential energy of MgH then they will raise their centre of mass a height of:

H=U22g

If the Olympic champion can reach 9 ms−1 launch speed then since the acceleration due to gravity is g=10 ms−2 we expect him to be able to raise his centre of gravity height of H=4 metres. If he started with his centre of gravity about 1.5 metres above the ground and made it pass 0.5 metres below the bar then he would be expected to clear a bar height of 1.5+4+0.5=6 metres. In fact, the American champion Tim Mack won the Athens Olympic Gold medal with a vault of 5.95 metres (or 19′614" in feet and inches) and had three very close failures at 6 metres, knowing he had already won the Gold Medal, so our very simple estimates turn out to be surprisingly accurate.

John D. Barrow is Professor of Mathematical Sciences and Director of the Millennium Mathematics Project at Cambridge University.

3 0
2 years ago
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