Ask question

Ask question

All categories

3 years ago

6

PLEASE HELP... Flight 4358 took off from the DFW airport at an elevation of 48°. If the airplane has been traveling for 1200 ft.

, approximately how far above the ground is the airplane?
1,614 ft.

1,793 ft.

802 ft.

891 ft.

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

I took the test and it's 891 ft. :)

You might be interested in

What does it mean if the probability is 40% that a spinner will stop in a red section? Question 5 options: A. There is one chanc

Svetllana [295]

C would be the obvious solution.

4 0

3 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST !!

zimovet [89]

Answer: #1 is the answer. Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line. Notice how the pairs of alternating exterior angles lie on opposite sides of the transversal but outside the two parallel lines.

Step-by-step explanation: Parallel Lines: Definition: We say that two lines (on the same plane) are parallel to each other if they never intersect each other, ragardless of how far they are extended on either side. Pictorially, parallel lines run along each other like the tracks of a train.

7 0

3 years ago

Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago

Which property is illustrated by the equation 3 (m n) = (3 m) n?

Sav [38]

Associative property

8 0

3 years ago

Read 2 more answers

Segment AE is a median for Triangle ABC.

Shalnov [3]

Step-by-step explanation:

$BC = BE + CE...(\because B-E-C) \\ \\ \therefore \: BC = 3x - 5 + x + 1 \\ \\ \therefore \: BC = 4x - 4 \\ \\ \huge \red{ \boxed{ \therefore \: BC = 4(x - 1)}}$

4 0

3 years ago