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strojnjashka [21]
3 years ago
10

How are dot products useful?

Mathematics
2 answers:
nata0808 [166]3 years ago
6 0

Answer:

An important use of the dot product is to test whether or not two vectors are orthogonal. Two vectors are orthogonal if the angle between them is 90 degrees. ... Thus, two non-zero vectors have dot product zero if and only if they are orthogonal.

Step-by-step explanation:

Alex777 [14]3 years ago
3 0
Dot product is to test whether or not two vectors are orthogonal. Two vectors are orthogonal if the angle between them is 90 degrees. ... Thus, two non-zero vectors have dot product zero if and only if they are orthogonal.
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15+c=17.50 First you subtract 15 by 15 because you switch operations from adding to subtracting then you would put C under it Second subtract 17.50 and 15 and that is 2.50 so C= 2.50

i need help please What is the total number of drawSprites(); you can have in a program? *

Step-by-step explanation:

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A plane intersects both nappes of a double-napped cone but does not go through the vertex of the cone. What conic section is for
lisabon 2012 [21]
<span>When a plane intersects both nappes of a double-napped cone but does not go through the vertex of the cone, the conic section that is formed by the intersection is a curve known as hyperbola. 
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3 0
3 years ago
The total cost of renting a vacation
sleet_krkn [62]

Answer:

y=125x+75

Step-by-step explanation:

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700=125(5)+B

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4 0
3 years ago
The antibiotic clarithromycin is eliminated from the body according to the formula A(t) = 500e−0.1386t, where A is the amount re
PolarNik [594]

Answer:

Time(t) = 11.61 hours (Rounded to two decimal place)

Step-by-step explanation:

Given: The antibiotic  clarithromycin is eliminated from the body according to the formula:

A(t) = 500e^{-0.1386t}                 ......[1]

where;

A - Amount remaining in the body(in milligram)

t - time in hours after the drug reaches peak concentration.

Given: Amount of drug in the body is reduced to 100 milligrams.

then,

Substitute the value of A = 100 milligrams in [1] we get;

100= 500e^{-0.1386t}

Divide both sides by 500 we get;

\frac{100}{500}=\frac{ 500e^{-0.1386t}}{500}

Simplify:

\frac{1}{5} = e^{-0.1386t}

Taking logarithm both sides with base e, then we have;

\log_e (\frac{1}{5})= \log_e (e^{-0.1386t})

\log_e (\frac{1}{5})=-0.1386t         [ Using \log_e e^a =a ]

or

\log_e (0.2)=-0.1386t

-1.6094379124341 = -0.1386t

 [using value of \log_e (0.2) = -1.6094379124341 ]

then;

t = \frac{-1.6094379124341}{-0.1386}

Simplify:

t ≈11.61 hours.

Therefore, the time 11.61 hours(Rounded two decimal place) will pass before the amount of drug in the body is reduced to 100 milligrams


6 0
3 years ago
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