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ziro4ka [17]
3 years ago
7

PLZ HELP MEEEE, will give brainliest

Mathematics
1 answer:
Gwar [14]3 years ago
8 0

Answer:

It is either B or D but i would say B

Step-by-step explanation:

You might be interested in
BD=16 and ÁT is the<br> perpendicular bisector of BD. Y=?
iren2701 [21]

Answer:

y= 5

Step-by-step explanation:

Given

BD = 16

BC = 2y - 2

CD = y + 3

Required

Find y

BD = BC + CD

So, the expression becomes

16 = 2y - 2 + y + 3

Collect like terms

16 = 2y  + y- 2 + 3

16 = 3y+1

Collect like terms

3y = 16 - 1

3y = 15

Make y the subject

y= 15/3

y= 5

8 0
3 years ago
Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect
Colt1911 [192]

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid  

z = x2 + y2 when the coordinates (x,y,z)=(cos(πt), sin(πt), t) of the helix satisfy the equation of the paraboloid. That is, when

\bf (cos(\pi t), sin(\pi t), t)

But  

\bf cos^2(\pi t)+sin^2(\pi t)=1

so, the helix intersects the paraboloid when t=1. This is the point

(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The <em>tangent vector</em> to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface  

\bf z=x^2+y^2

at the point (-1, 0, 1) is given by

\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)

where

\bf f(x,y)=x^2+y^2

since

\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y

so, a normal vector to the tangent plane is

(-2,0,-1)

Hence, <em>a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane </em>is given by

\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.  

But we now

\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta

where  

\bf \theta= angle between the tangent vector and its projection onto the tangent plane. So

\bf \pi^2=(\sqrt{4+\pi^2}\sqrt{\pi^2+1})cos\theta\rightarrow cos\theta=\frac{\pi^2}{\sqrt{4+\pi^2}\sqrt{\pi^2+1}}=0.8038

and

\bf \theta=arccos(0.8038)=0.6371\;radians

7 0
3 years ago
A.$454.75<br> B.$502.75<br> C.$327.75<br> D.$327.25<br> E.None of These
ehidna [41]

Answer:

D. 327.25

Step-by-step explanation:

You start out with a gross pay of $415.00. When you have a deduction from your gross pay, it means it gets subtracted. So subtract the deductions of 48, 31.25, and 8.50 from 415 and it leaves you with 327.25.

415 48-31.25-8.50 = 327.25

3 0
3 years ago
What does 2 six equals
RUDIKE [14]
12 because 6+6=12 or 6 times 2 = 12
3 0
3 years ago
Read 2 more answers
How do you find JK?
Naily [24]

Answer:

JK = 13

Step-by-step explanation:

Note that corresponding sides are congruent.

Thus ST = KL and JK = RS

Given ST = 16RS and KL = 13JK + 39, thus

KL = 13JK + 39 ( KL = ST = 16RS ), substituting

16RS = 13RS + 39 ( JK = RS )

Subtract 13RS from both sides

3RS = 39 ( divide both sides by 3 )

RS = 13

However, RS and JK are corresponding sides and congruent, thus

JK = RS = 13

4 0
4 years ago
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