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2 years ago

Five times a number is 5 more than the number

2 answers:

5 0

The answer to your question is.... 5X+5!!!
Five times a number would express 5X and 5 more would express +5!!

I hope this has helped!

sattari [20]2 years ago

4 0

5x+5=

Hope this helped

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Jane moved into a flat with two other students. the rent was £540 per month altogether and the council tax worked at £216 each p

Delvig [45]

9873 - 9586 = 287 units used cost is 287 * 27.2 = 7806.4 p= 78.06 + standing charge of 12.40

total gas 80.46 Jayne's share 26.16 (because they have to pay part of a penny, so round up) 
electricity 
3542 - 3187 = 355 units used cost 355 * 7.1 2520.5 p = 25.21 + standing charge of 9.10 

total electricity 34.31 Jayne's share 11.44 (because they have to pay part of a penny, so round up) 
Rent * 3 months = 540 * 3 = 1620 Jayne's share 1620 / 3 = 540 

total cost 3 months for Jayne 
540 + 11.44 + 26.16+ 54 = 631 pounds 60 pence

4 0

3 years ago

A ball is thrown in the air from a platform that is 96 feet above ground level with an initial vertical velocity of 32 feet per

pishuonlain [190]

Answer:

y = -16 (x - 1)^2 + 112

The object lands on the ground in approximately 3.6s

Explanation:

The equation given is that of a parabola.

Now the maximum (local) point of a parabola is the vertex. Therefore, if we want to rewrite our function in the form that would be used to find the maximum height, then that form must be the vertex form of a parabola.

The vertex form of a parabola is

$y=a(t-h)^2+k$

where (h, k) is the vertex.

The only question is, what is the vertex for our function h(t)?

Remember that if we have an equation of the form

$y=ax^2+bx+c$

then the x-coordinate of the vertex is

$h=-\frac{b}{2a}$

Now in our case b = 32 and a = -16; therefore,

$h=\frac{-32}{2(16)}=1$

We've found the value of the x-coordinate of the vertex. What about the y-coordinate? To get the y-coordinate, we put x = 1 into h(t) and get

$k=-16(1)+32(1)+96=112$

Hence, the y-coordindate is k = 112.

Therefore, the vertex of the parabola is (1, 112).

With the coordinates of the vertex in hand, we now write the equation of the parabola in vertex form.

$h(t)=a(t-1)^2+112$

The only problem is that we don't know what the value of a is. How do we find a?

Note that the point (0, 96) lies on the parabola. In other words,

$h(0)=-16(0)^2+32(0)+96=96$

Therefore, the vertex form of the parabola must also contain the point (0, 96).

Putting in t = 0, h = 96 into the vertex form gives

$96=a(0-1)^2+112$ $96=a+112$

subtracting 112 from both sides gives

$a=-16$

With the value of a in hand, we can finally write the equation of the parabola on vertex form.

$\boxed{h\mleft(t\mright)=-16\left(t-1\right)^2+112.}$

Now when does the object hit the ground? In other words, for what value of t is h(t) = 0? To find out we just have to solve the following for t.

$h(t)=0.$

We could either use h(t) = -16t^2 + 32t + 96 or the h(t) = -16(t - 1)^2 + 112 for the above equation. But it turns out, the vertex form is more convenient.

Thus we solve,