Alex purchased a motorbike that costs $3,500. He’ll pay for the motorbike through an installment plan. The installment plan requ

Answer:

a) The first inequality is 100 + 55x > 150 + 51x;

b) The final inequality results in x > 12.5

c) Sal's mother will need to use the second phone for at least 13 months.

Step-by-step explanation:

a) Let x represent the number of months.

1. The first phone is priced at $100, with a monthly fee of $55 for unlimited use, leading to a total cost of $(100 + 55x) for x months.

2. The second phone costs $150 with a monthly fee of $51 for unlimited use, resulting in a total of $(150 + 51x) for x months.

3. For the second phone to be cheaper, we set up the inequality:

150 + 51x < 100 + 55x

which simplifies to

100 + 55x > 150 + 51x

b) Now solve this:

55x - 51x > 150 - 100

4x > 50

so x > 12.5

c) This means Sal's mother has to retain the second phone for at least 13 months (since x > 12.5).

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

The acceleration is:

a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

<pthus our="" equation="" is:="">

h(x) = a*x^2 + v0*x + h0.

<pexamining the="" table:="">

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

            h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

<pthus our="" resulting="" equation="" reads:="">

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) =  (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>

2G=B+3
G=B-4

2(B-4)=B+3
2B-8=B+3
B=11

Bert possesses 11 cents

This can be approached using trigonometry. The adjacent side corresponds to the length of the shadow, while the unknown represents the opposite side, which is Chad's height.

Tan (a) = opposite / adjacent

Opposite = tan(23) * 14.3 ft

Thus, the opposite side equals 6.1 ft, which is Chad's height