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2 years ago

The plane flew over a beacon which is 280 miles from bristol. write down the gmt times when the plane flew over the beacon.

Mathematics

1 answer:

eimsori [14]2 years ago

7 0

According to graph, GMT time when the plane flew over the beacon was 9:24.

While looking closely at the graph on x axis , one will notice that each hour is divided into 5 parts and hence each part has 12 minutes. looking at y axis, one may find that each part on y axis resembles 20miles. Hence at 280 miles away from Bristol, The GMT time is 9:24.

<h3>What is a graph?</h3>

In mathematics, a graph is a visual representation or diagram that shows facts or values in an ordered way.

The relationships between two or more items are frequently represented by the points on a graph.
The study of graphs, which are mathematical constructions used to represent pairwise relationships between things, is known as graph theory in mathematics. Vertices (also known as nodes or points) make up a graph in this sense, which are connected by edges (also called links or lines).
Undirected graphs, in which edges connect two vertices symmetrically, are distinguished from directed graphs, in which edges connect two vertices asymmetrically. One of the main topics of study in discrete mathematics is graphs.

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Answer:

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3 years ago

The length of a rectangle is 24 units. Can the perimeter x of the rectangle be 60 units when its width y is 11 units?

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4 0

4 years ago

Read 2 more answers

Use these functions to answer the questions.

Elenna [48]

Part A
We have $f\left(x\right)=\left(x+3\right)^2-1$ . To solve for the x-intercept, we set f(x) equal to 0. That is
$\left(x+3\right)^2-1=0$

$\left(x+3\right)^2=1$

Take the square root of both sides,
$x+3=1$

$x=-2$

The x-intercept is (-2,0).

To solve for the y-intercept, we set x=0. That is
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The y-intercept is (0, 8)

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have $g\left(x\right)=-2x^2+8x+3$ .
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$=-2\left(x^2-4x\right)+3$

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$=-2\left(x^2-4x+4\right)+3-2\left(4\right)$

Simplify
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Part C
We have $h\left(t\right)=-16t^2+28t$ .
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