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3 years ago

Choose the correct graph to fit the inequality x^2 - y^2 < 9

1 answer:

5 0

There is no pictures to choose the right one among tem.

However the equation could be rewritten as follows:

y=√(x²-9)

It is a HORIZONTAL parabola with the following constraint: 3=<x=<-3

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Nathaniel builds birds and birdhouses using Lego blocks. Let B represent the number of birds and H represent the number of birdh

N76 [4]

Answer:

At-most 3 birdhouses.

Step-by-step explanation:

Let B represent the number of birds and H represent the number of birdhouses that Nathaniel can build with his Lego blocks.

We have been given that Nathaniel builds birds and birdhouses using Lego blocks. The inequality $43B+215H\leq3000$ represents the number of birds and birdhouses Nathaniel could build.

To find the number of birdhouses that Nathaniel can build after making 50 birds, we will substitute B=50 in our given inequality and then solve for H.

$43*50+215H\leq3000$

$2150+215H\leq3000$

Let us subtract 2150 from both sides of our inequality.

$2150-2150+215H\leq3000-2150$

$215H\leq850$

Let us divide both both sides of our inequality by 215.

$\frac{215H}{215}\leq\frac{850}{215}$

$H\leq3.953488$

As Nathaniel can make build less than or equal to 3.953488 and the biggest integer less than or equal to 3.953 is 3, therefore, Nathaniel can build at-most 3 birdhouses with remaining Lego blocks.

3 0

2 years ago

Read 2 more answers

Need help ASAP! due soon!!!

-BARSIC- [3]

Answer:

Option B

Step-by-step explanation:

First make x = 4 and y = 3 in the equation y = -1/2x + 5

it should look like 3 = -1/2(4) + 5 now we can solve

-1/2 * 4 = -2 → 3 = -2 +5

-2 + 5 = 3 → 3 = 3

5 0

2 years ago

Find the Volume: 3.2 cm<br> 7 cm<br> 4.8 cm

REY [17]

V = 0.5*b*h*l
v = 0.5*4.8*3.2*7
v = 0.5*107.52
v = 53.76cm3

7 0

2 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

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

6 0

3 years ago

Please help me. This is due today

sattari [20]

Answer:

it answer is 3 because it is trigonometry

3 0

2 years ago