Kathy and Mark Smith believe investing in their retirement is critical. Kathy began investing 20% of each paycheck in a retireme

12 m^2/h to cm^2/min

12 m^2/h × 1 h/60 min = 0.2 m^2/min

0.2 m^2/min × 10000 cm^2/1 m^2 = 2000 cm^2/min

2000 cm^2/min

Answer:

Step-by-step explanation:

The triangle provided is a right triangle. The length of the opposite side is 244 yards, and the adjacent side measures 119 yards. To find the hypotenuse, we would utilize the Pythagorean theorem. The formula can be written as

Hypotenuse^2 = opposite side^2 + adjacent side^2

Thus, Hypotenuse^2 = 244^2 + 119^2

Hypotenuse^2 = 59536 + 14161

Leading to Hypotenuse = √73697 = 271.5 yards.

cos θ = adjacent side/hypotenuse

cos θ = 119/271.5 = 0.4383

Calculating θ gives us θ = Cos^-1(0.4383) = 64 degrees

The remaining angle is

90 - 64 = 26 degrees

pete 23 autos alex 23 alex agrega 16 autos luego 23 más 23 son 46

I believe the answer is 30.51, but you should verify by dividing.

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>