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Oksi-84 [34.3K]
3 years ago
12

Graph the function. Upload a handwritten copy of your graph as your final answer.

Mathematics
1 answer:
Serhud [2]3 years ago
7 0
The function is divided in 2 parts:

For x<0, the function is y=x, which is a linear function, and its graph is a line.

For x≥0, the function is y=-x^2, which is a quadratic function, and its graph is a parabola.


Note that y=x is the line containing the points (0, 0), (1, 1), (5, 5) etc, that is, it is the line passing through the origin, with an inclination of 45° with the positive x-axis.


The parabola y=-x^2 is the parabola y=x^2 reflected with respect to the x-axis. It contains points like (0, 0), (-1, 1), (2, -4) etc...

The 2 graphs, with domain all real numbers, are shown in the first picture.

The second picture is the graph of the function described in our problem. We use the first picture, but restrict the domains.

That is, we delete the part of the line corresponding to x≥0, and the part of the parabola corresponding to x<0.

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Hoi!

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A building is 3 ft from an 8​-ft fence that surrounds the property. A worker wants to wash a window in the building 13 ft from t
zloy xaker [14]

Answer:

Length of the ladder used by worker = 17 feet

Step-by-step explanation:

Given:

Height of the window from the ground = 13 ft

Distance of fence from the building = 3 ft

Distance of ladder from the building = (3+8) = 11 ft

We have to find the length of the ladder.

Let the length of the ladder be 'x'

From the diagram we can also say that 'x' is the hypotenuse of the right angled triangle.

Using Pythagoras formula:

⇒ hypotenuse\ 'x' =\sqrt{(perpendicular)^2+(base)^2}

Here base length = 11 ft

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Plugging the values:

⇒ x=\sqrt{(13)^2+(11)^2}

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6 0
4 years ago
The first term of a geometric sequence is 32, and the 5th term of the sequence is 818 .
sammy [17]

Answer:

32,24,18,\frac{27}{2} ,\frac{81}{8}

Step-by-step explanation:

Let x, y , and z be the numbers.

Then the geometric sequence is 32,x,y,z,\frac{81}{8}

Recall that  term of a geometric sequence  are generally in the form:

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This implies that:

a=32 and ar^4=\frac{81}{8}

Substitute a=32 and solve for r.

32r^3=\frac{81}{8}

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x² + 2x - 2x - 4

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