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tangare [24]
3 years ago
12

Sam and Lou need a total of 1 foot of wire for a science project. Sam's wire measured 7/8- foot long. Do they have enough wire f

or science project? Explain your reasoning
Mathematics
1 answer:
Reil [10]3 years ago
5 0

Answer:

Yes, Sam and Lou have enough wire for science project.

Step-by-step explanation:

<u><em>There is some data missing in the question, so complete question is below:</em></u>

Sam\ and\ Lou\ need\ a\ total\ of\ 1\ foot\ of\ wire\ for\ science\ project.Sam's\ wire\ measured\ 8/12\ foot\ long.\ Lou's\ measured\ 7/8-\ foot \ \ long.Do\ they\ have\ enough\ wire\ for\ the\ science\ project?Explain \ your\ reasoning.

Now, to get that whether they have enough wire for the science project or not.

Measurement of Sam's wire = \frac{8}{12} \ foot.

Measurement of Lou's wire = \frac{7}{8}\ foot.

Now, adding both the wires of Sam and Lou:

\frac{8}{12} +\frac{7}{8} \\\\=\frac{16+21}{24} \\\\=\frac{37}{24}

=1\frac{13}{24} \ feet.

<em>As, Sam and Lou need a total of 1 foot of wire for the science project.</em>

<em>And, we see that the Sam's wire and Lou's wire combining gets </em>1\frac{13}{24}\ feet<em> which is</em> <em>more than 1 foot.</em>

<em>Thus, wire they have is enough for science project.</em>

Therefore, yes, Sam and Lou have enough wire for science project.

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Step-by-step explanation:

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2 years ago
HELP PLEAAASEEE Ireland opened a bag or skittles and recorded the colors and their frequencies.
Anna11 [10]

The correct answer is 25%

<h3>What is Relative Frequency?</h3>
  • The number of times an event occurs divided by the total number of events occurring in a given data.

<h3>How to solve the problem?</h3>
  • This problem can be solved by following steps.
  • The data of color and frequency is given in table.
  • We need find the relative frequency for Purple

First calculate the the total number of frequency

18+20+10+16

= 64

The total number of frequency is 64

Hence the relative frequency of purple is 16/64

Therefore , relative frequency of purple is 1/4 = 0.25

Therefore 25% of purple candies are there

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2 years ago
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Graph -2x-3 please!!!!
stich3 [128]

Answer:

Step-by-step explanation:

4 0
3 years ago
17. Identify the amplitude, period, and vertical shift for the
Stella [2.4K]

Answer:

f(x)=2cos(x+\frac{\pi}{2})+3

Step-by-step explanation:

Because the function is symmetric about the y-axis, using the cosine function is most appropriate.

<u>Refer to the equation for a cosine function:</u>

<u />f(x)=acos(bx+c)+d<u />

Amplitude: |a|

Period: \frac{2\pi}{|b|}

Phase shift: -\frac{c}{b}

Midline: y=d

The amplitude would be the average of the maximum and minimum y-values of the function, which would be |a|=|\frac{5-1}{2}| =|\frac{4}{2}| =|2|=2.

The value of b in \frac{2\pi}{|b|} represents the length of the period, so since the length of the period is 2\pi, this means that b=\frac{2\pi}{|2\pi|} =1.

The phase shift, -\frac{c}{b}, describes the horizontal shift of a function. Because the phase shift is -\frac{\pi}{2}, then we can set up the equation -\frac{\pi}{2}=-\frac{c}{1} where we determine c=\frac{\pi}{2}.

The midline (or vertical shift), d, is the horizontal line that passes through between the maximum and minimum points, which the function oscillates. In this case, the midline would be located at the line y=3, therefore, d=3.

Putting all our information together, your final equation is:

f(x)=2cos(x+\frac{\pi}{2})+3

3 0
3 years ago
A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500
timurjin [86]
<span>4.98 ft/s Let's determine the distance between the man and the woman for the moment that she's been walking 15 minutes. For this you can create a right triangle where one leg is 500 ft long (the east west difference between their locations) and the other leg is (distance man walked for 20 minutes + distance woman walked for 15 minutes). So Distance man walked = 20 min * 60 s/min * 2 ft/s = 2400 ft. Distance woman walked = 15 min * 60 s/min * 3 ft/s = 2700 ft. So the north south different in the man and woman's location is 2400+2700 = 5100 ft and will be increasing by 5 ft/sec. Creating a function of time (in seconds) for the distance the two people are apart is f(t) = sqrt(500^2 + (5100 + 5t)^2) where t = number of seconds from the 15 minutes the woman has been walking. For rate of change, you want the first derivative of the function. So let's calculate it. f(t) = sqrt(500^2 + (5100 + 5t)^2) f(t) = sqrt((5100 + 5t)^2 + 250000) f'(t) = d/dt[ sqrt((5100 + 5t)^2 + 250000) ] f'(t) = 0.5((5t + 5100)^2 + 250000)^(-0.5) * d/dt[ (5t + 5100)^2 + 250000 ] f'(t) = d/dt[ (5t + 5100)^2 ] / (2 * sqrt((5t + 5100)^2 + 250000)) f'(t) = 2(5t + 5100) * d/dt[ 5x + 5100 ]/(2 * sqrt((5t + 5100)^2 + 250000)) f'(t) = 5(5t + 5100/sqrt((5t + 5100)^2 + 250000) f'(t) = (25t + 25500)/sqrt((5t + 5100)^2 + 250000) Now calculate f'(t) for t = 0. So f'(t) = (25t + 25500)/sqrt((5t + 5100)^2 + 250000) f'(0) = (25*0 + 25500)/sqrt((5*0 + 5100)^2 + 250000) f'(0) = 25500/sqrt((5100)^2 + 250000) f'(0) = 25500/sqrt(26010000 + 250000) f'(0) = 25500/sqrt(26260000) f'(0) = 25500/5124.45119 f'(0) = 4.976142626 ft/sec So the man and woman are moving away from each other at the rate of 4.98 ft/s.</span>
8 0
4 years ago
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