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3 years ago

MATHS HELP PLEASE!!!! Can anyone help me with question 1????

Mathematics

2 answers:

dem82 [27]3 years ago

7 0

Each centimeter is three people so you just have to count the centimeter and multiply that by three to get the answer
Math 9 friends
History 6 friends
Eng/sci 15 friends

torisob [31]3 years ago

3 0

Answer: For part A, Yakob has 9 friends that like Maths best, 6 friends that like History the best, & 15 friends that prefer <em>either</em> English or Science the best.

Part B: There's a graph I made below.

Step-by-step explanation:

You have to use proportions to answer this question. The bar graph is 10 cm total, therefore it is proportional to the 30 friends that Yakob has in total. After that, you can make proportions out of the data in the chart. For example, History is 2 cm out of the 10 cm total. Think proportions! You had to multiply 3 to 10 to get 30, so you have to apply the same thing to the numerator. In this case, multiply 3 to 2 and you'll get 6. 2/10=6/30! I repeated that for the rest and put it into a graph. The prompt wanted you to combine English and Science, so I did that after separately making their proportions. Hope this helps!

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Then x = 2 or 8
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3 years ago

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

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Answer:

Option A. 0.20

Step-by-step explanation:

Observations from the given frequency table are :

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= 15

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Then, probability that the next bed sold will be a twin bed,

$P=\frac{\text{favorable outcome}}{\text{total number of possible outcomes}}$

$=\frac{3}{15}$

= 0.20

Therefore, the probability that the next bed sold is a twin bed will be 0.20 or 20%.

Option A. is the correct answer.

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AleksAgata [21]

-1^2/8 = 1/8 is the answer

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