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Snowcat
3 months ago
15

John runs 500 feet in 1 minute. Identify the correct conversion factor setup required to compute John's speed in inches per seco

nd.
Mathematics
2 answers:
PIT_PIT [12.4K]3 months ago
7 0

Answer:

refer to the method

Step-by-step explanation:

Keep in mind that

1\ ft=12\ in

To change feet to inches, multiply by 12

1\ min=60\ sec

To convert minutes into seconds, multiply by 60

we have

500\frac{ft}{min}=500(\frac{12}{60})=500(\frac{1}{5})=100\frac{in}{sec}

Inessa [12.5K]3 months ago
5 0

Answer:

\frac{500\text{ ft}}{\text{ 1 min}}\times \frac{1 \text{ min}}{\text{ 60 sec}}\times \frac{12\text{ inches}}{\text{1 ft}}

Step-by-step explanation:

We know John runs 500 feet within 1 minute. We need to identify the correct setup of conversion factors needed to calculate John's speed in inches per second.

It is established that 1 minute equals 60 seconds and 1 foot is equivalent to 12 inches.

\frac{500\text{ ft}}{\text{ 1 min}}\times \frac{1 \text{ min}}{\text{ 60 sec}}\times \frac{12\text{ inches}}{\text{1 ft}}

Thus, our necessary conversion factor setup would be \frac{500\text{ ft}}{\text{ 1 min}}\times \frac{1 \text{ min}}{\text{ 60 sec}}\times \frac{12\text{ inches}}{\text{1 ft}}.

Now, let's convert 500 feet in 1 minute into inches per second:

\frac{500}{1}\times \frac{1}{\text{ 60 sec}}\times \frac{12\text{ inches}}{1}

\frac{500}{1}\times \frac{1}{\text{ 5 sec}}\times \frac{1\text{ inches}}{1}

\frac{500\text{ inches}}{\text{ 5 sec}}

\frac{100\text{ inches}}{\text{ sec}}

Thus, John runs at a speed of 100 inches per second.

   

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AnnZ [12381]

Answer:

The distance from the point (0,1,1) to the specified line is zero.

Step-by-step explanation:

Considering the parametric equations of the line,

x=2t, y=5-2t, z=1+t

In order to calculate the distance from (0,1,1), we must remove t from the equations above, such that

y=5-2t=5-x\implies x+y=5=4+1=4+z-t=4+z-\frac{1}{2}x

\implies 3x+2y-2z-8=0\hfill (1)

whose direction ratios are (l,m,n)=(3,2,-2) and the distance from point (a,b,c)=(0,1,1) is defined as

\frac{al+bm+cn}{\sqrt{l62+m^2+n^2}}=\frac{(3\times 0)+(2\times 1)+(-2)(1)}{\sqrt{3^2+2^2+(-2)^2}}=\frac{0+2-2}{\sqrt{17}}=0

The distance between the point (0,1,1) and (1) amounts to zero. Therefore, the point (0,1,1) is located on the line (1).

8 0
3 months ago
The width of a rectangle is 61 centimeters more than the length. The perimeter is 406 centimeters. Find the length and the width
lawyer [12517]

Step \; 1: \; Assign \; Variables \; for \; the \; unknown \; that \; we \; need \; to \; find

Let \; x \; be \; length \; of \; the \; rectangle

Step \; 2: \; Set \; up \; equation \; based \; on \; information \;\\ given \; about \; the \; rectangle

Statement \; 1: Width \; of \; a \; rectangle \; \\is \; 61cm \; more \; than \; the \; length\\\\Width \; = \; 61+x\\\\Statement \; 2: \; The \; perimeter \; is \; 406cm\\\\Perimeter=2(Length+Width)\\Perimeter =2(x+61+x)\\\\So \; the \; mathematical \; equation \; would \; be \\ 2(x+61+x)=406

Step \; 3: \; Solve \; the \; equation \; by \\ undoing \; whatever \; is \; done \; x.\\\\2(x+61+x)=406\\Group \; and \; Combine \; like \; terms \; inside \; the \; parenthesis\\\\2(2x+61)=406\\Distribute \; 2 \; in \; the \; left \; side \; of \; the \; equation\\\\4x+122=406\\Subtract \; 122 \; on \; both \; sides\\\\4x+122-122=406-122\\Simplify \; on \; both \; sides\\\\4x=284\\Divide \; on \; both \; sides\\\\\frac{4x}{4}=\frac{284}{4}\\Simplify \; fractions \; on \; both \; sides\\\\x=71

Conclusion:\\Length=x=71cm\\Substituting \; 71 \; for \; x \; and \; find \; Width \; value.\\Width=61+x=71+61=132cm\\\\Length \; is \; 71 cm \; and \; Width \; is 132cm

5 0
2 months ago
Tanmay had some chocolates with him. He gave one-third of them to Akash and one-fifth of them to Sharad. Tanmay could do so with
babunello [11817]

Answer:

Possible values for X include;

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Step-by-step explanation:

The parameters provided are

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Consequently, since both 3 and 5 divide X,

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Thus, the values of X based on this minimum factor are as follows;

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Therefore, potential values for X form an arithmetic series: a + (n - 1) × d

Where:

a = 15

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This results in;

15, 30, 45, 60

7 0
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If only two points are needed to determine a​ line, why are three ordered pair solutions or points found for a linear​ equation?
zzz [12365]

To plot the graph of a linear equation, two points suffice to create a line.

Nevertheless, errors can occur; thus, when the line is drawn, it may seem accurate. Two points will always define a line, regardless of the correctness of those points.

To confirm that the points are accurate, we find a third point to verify that the two previously calculated points are correct and that all points align on the same line. If they don’t, an error has been made.

Option A) is the correct choice

7 0
3 months ago
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