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Gwar [14]
3 years ago
7

Now, expand the expression 1/2(13x+5x-6-8x+14)=1/2(10x+8)

Mathematics
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

5x + 4

Step-by-step explanation:

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I really need help with this problem?
trasher [3.6K]

7 cm, it’s a step by step equation

3 0
3 years ago
58-(1/4)2 i dont get how to solve this
xeze [42]

Answer:

57.5

Step-by-step explanation:

First you would have to multiply following PEMDAS so

(1/4)(2)

=0.5

Then subtract 58 and 0.5

58-0.5

=57.5

and thats your answer

5 0
3 years ago
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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
3 years ago
HELP PLEASE pre-calc
nikitadnepr [17]
the second choice, and the last choice are correct.

the first one, x can be any number.
the second one, x can be any number but not -1(the denominator cannot be 0)
the third one, x has to be ≥ 1/2, because 2x-1 cannot be negative.
the fourth one, x can be any number but not -1 or -4
the last one, x can be any number but not -3
5 0
3 years ago
Simplify the expression below.
Mazyrski [523]

Answer:

Its either C or D

good luck

have a good day - be safe - and healthy too

4 0
3 years ago
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