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3 years ago

How do you solve this?

Mathematics

1 answer:

SpyIntel [72]3 years ago

4 0

I am not a college student so I am sorry or you could ask your friends.

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Jose rented a truck for one day. There was a bose fee of $19.95, and there was an additional charge of 84 cents for each mile dr

nevsk [136]

Answer:

Step-by-step explanation:

Let the number of miles = x

C ost = Base + 0.84x

Cost = 220.71 dollars

Base = 19.95

220.71 = 19.95 + 0.84x Subtract 19.95 from both sides

220,71 - 19.95 = 0.84 x Combine

200.76 = 0.84x Divide by 0,84

200.76/0.84 = x

x = 239 miles.

Remark: The trick to this question is to make sure that total Cost, the Base amount, and cost / mile are all in dollars. Don't mix and match.

5 0

3 years ago

The local hardware store has blue buckets that hold 2 gallons of water and white buckets that hold 5 gallons of water. You bough

Katarina [22]

Answer:

4 White, 3 Blue

Step-by-step explanation:

5w+2b=26

Closest five multiple that leaves an even remainder is 4.

5×4=20

26-20=6÷2=3

5 0

3 years ago

Read 2 more answers

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

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

6 0

3 years ago

What percent of 55 is 34

djyliett [7]

answer

.618 18 repeats

Step-by-step explanation:

4 0

3 years ago

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Mr. Johnson sells erasers for $3 each. He sold 96 erasers last week and he sold 204 erasers this week

IgorC [24]

Answer:

288

Step-by-step explanation:

it's pretty simple, if u do 96 * 3, it will be that. yw :)

4 0

3 years ago

Read 2 more answers