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3 years ago

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A company sells cereal in two different-sized boxes: a 17-ounce box for $2.89, and a 24-ounce box for $3.29. How do the unit pri

ces of the two boxes compare?

1 answer:

son4ous [18]3 years ago

4 0

The 17 ounce cereal is $0.17 per ounce and the 24 ounce cereal is $0.13 per ounce.

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How to solve this? Please show work.<br><br><br> x+3 /4 = x-1 /5

Vinil7 [7]

This will take me a long time, but... $x+ \frac{3}{4} = x- \frac{1}{5} =x+ \frac{3}{4} -x = x- \frac{1}{5} -x = \frac{3}{4} =\frac{1}{5} =\frac{3}{4} - \frac{3}{4} =\frac{1}{5} - \frac{3}{4}
$ $=\frac{-19}{20}$

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3 years ago

Alyssa is looking at a car. If she buys the car, she will pay $750 when she picks up the car (a down payment) and pay $314 each

soldier1979 [14.2K]

Answer:

a very expensive waste of money.

Step-by-step explanation:

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3 years ago

1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is

sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by $d(t)=\frac{t^{2}}{6}+4t$ where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

$V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}$

so let's take the first interval:

(a) [1.95, 2.05]

$V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}$

we get that:

$d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125$

$d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167$

so:

$V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s$

(b) [1.995, 2.005]

$V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}$

we get that:

$d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831$

$d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835$

so:

$V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s$

(c) [1.9995, 2.0005]

$V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}$

we get that:

$d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358$

$d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358$

so:

$V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s$

(d) [2, 2.00001]

$V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}$

we get that:

$d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333$

$d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333$

so:

$V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s$

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

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3 years ago

Convert 36 yards into inches

andrey2020 [161]

Answer:

36yd = 108ft

108ft = 1296

Step-by-step explanation:

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3 years ago

Consider two quantities u and v that are related by the expression vp=cuq, where c is a constant. the exponents p and q are not

Mazyrski [523]

Answer:

y = (q/p)x +log(c)/p

Step-by-step explanation:

$v^{p}=cu^{q}\qquad\text{original equation}\\\\p\log{(v)}=\log{(c)}+q\log{(u)}\qquad\text{take the log}\\\\py=log{(c)}+qx\qquad\text{substitute x and y}\\\\y=\dfrac{\log{(c)}+qx}{p}$

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3 years ago