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2 years ago

15

What is 8 times 56 and what is 7 time 14

1 answer:

sp2606 [1]2 years ago

5 0

8*56=448 and 7*14=98

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586 x 10^6 written in standard notation

IgorLugansk [536]

Hello,

Shall we begin?

= 586 x 10^6
=-586 x 1,000,000
= 586 x 1,000,000
= 586,000,000

Answers: 586,000,000

7 0

2 years ago

a car dealership has 75 cars on the lot. Of the cars, 40% are black. how many black cars are on the lot?

My name is Ann [436]

Answer:

30

Step-by-step explanation:

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2 years ago

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GIVING BRAINIEST!!!!!

dexar [7]

Answer:

Clearly, the probability of drawing a heart out of the deck is 13/52, or 1/4. The probability of drawing an even number as card #2 (∈{2,4,6,8,10}) for any suit is 20/51, except when the first card drawn is an even heart, then the probability of the second draw is 19/51.

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2 years ago

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

const2013 [10]

Answer:

a.

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

b.

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

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

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

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

a.

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

b.

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

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

6 0

2 years ago

Miranda has 5/8 of a yard of ribbon. if she used 1/3 of it to make a hair bow, how much ribbon would she have left. PLEASE HELP

bezimeni [28]

Answer: 5/12

Step-by-step explanation:

Step 1

Whenever you have something OF something, you just mulltiply.

1/3 of 5/8=

1/3x5/8

5/24

She used 5/24, and started with 5/8

5/8-5/24= 5/12 left

8 0

1 year ago

Read 2 more answers