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3 years ago

Classify <1 and <3 as an angle of elevation or an angle of depression.

Mathematics

2 answers:

zhannawk [14.2K]3 years ago

5 0

Angle 1 is depression (it directly sinks) while 3 is elevation and still constant

weeeeeb [17]3 years ago

3 0

Angle 1 is an angle of depression because you start off looking directly horizontal and you look down to aim at the ground

Angle 3 is an angle of elevation. You start looking horizontally and then you look up

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Multi step problems what is -10(10+b)=5(3b+5)

Alla [95]

Answer:

b= -5

Step-by-step explanation:

-10(10+b)=5(3b+5)

Distribute

-100 -10b = 15b +25

Add 10b to each side

-100-10b+10b = 15b+10b +25

-100 = 25b+25

Subtract 25 from each side

-100-25 = 25b+25-25

-125 = 25b

Divide by 25

-125/25 =25b/25

-5 =b

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3 years ago

Dakota knows that EJ←→⊥CN←→ and CN←→∥KT←→ .

Lina20 [59]

Answer:

m∠CJV=90°
EV←→⊥KT←→

Step-by-step explanation:

Nothing in the problem statement tells you anything about the directions of lines LK or QT, so you cannot conclude they are parallel. You only know that EV crosses CN and KT at right angles.

EJ ⊥ CN means m∠CJV = 90°, as all angles at the intersection of EJ and CN are 90°.

6 0

3 years ago

A marching band stands in a triangle formation. The drum major stands at A, one of the vertices of the triangle. The whole forma

iragen [17]

Answer:

D. (1, 3)

Step-by-step explanation:

Based on the image uploaded, during the first march, move point A, 3 units left, the new position will (3, -4), then move point A again 5 units up, the new position will be (3,1).

From second march move point A 2 units left, the new position will be (1,1), finally move point A 2units up, the new position and last position will be (1, 3).

Thus, the drum major be standing on (1, 3)

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3 years ago

Since the opening up of the West, the US population has moved westward. To observe this, we look at the “population center” of t

Arlecino [84]

Answer:

attached below

Step-by-step explanation:

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3 years ago

Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.

Alika [10]

The Lagrangian for this function and the given constraints is

$L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)$

which has partial derivatives (set equal to 0) satisfying

$\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}$

This is a fairly standard linear system. Solving yields Lagrange multipliers of $\lambda_1=-\dfrac{32}{11}$ and $\lambda_2=-\dfrac{104}{11}$ , and at the same time we find only one critical point at $(x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ .

Check the Hessian for $f(x,y,z)$ , given by

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite, since $\mathbf v^\top\mathbf{Hv}>0$ for any vector $\mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top$ , which means $f(x,y,z)=x^2+y^2+z^2$ attains a minimum value of $\dfrac{480}{11}$ at $\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ . There is no maximum over the given constraints.

7 0

4 years ago