In 4 weeks, Mark earns an average of $400 per week. If, after working another 2 weeks at a constant salary, his average earnings

Answer:

5th-grade mean - 4.67

7th-grade mean - 3.46

5th-grade median - 5

7th-grade median - 3.5

Answer: A total descent of 8260 feet occurred for all four balloons.

Step-by-step explanation:

Based on the information provided:

Number of hot air balloons = 4

Minutes of descent = 7

Descent rate per minute = 295 feet

Thus, the overall change in altitude for all four balloons can be calculated as:

Thus, a total descent of 8260 feet was experienced by the four balloons.

(a) 4 <span>(b) y = sqrt(9 - (9/16)x^2) </span>The most accurate assumption for the equation based on the general format for an ellipse is: [[TAG_3]]x^2/16 + y^2/9 = 1 [[TAG_4]](a). An ellipse is symmetrical along both its major and minor axes. Thus, if you can calculate the area of the ellipse in one quadrant, multiplying that area by 4 will yield the total area of the ellipse, confirming the factor of 4 is accurate. [[TAG_5]] (b). The standard equation for an ellipse doesn't adequately represent a general function as it results in two y values for each x value. However, if we constrain ourselves to the positive square root, that issue can be resolved easily. Here’s how: [[TAG_6]] x^2/16 + y^2/9 = 1 [[TAG_7]] x^2/16 + y^2/9 - 1 = 0 [[TAG_8]] x^2/16 - 1 = - y^2/9 [[TAG_9]] -(9/16)x^2 + 9 = y^2 [[TAG_10]] 9 - (9/16)x^2 = y^2 [[TAG_11]] sqrt(9 - (9/16)x^2) = y [[TAG_12]] y = sqrt(9 - (9/16)x^2)

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>