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3 years ago

How many real solutions are there for the function graphed below?

Mathematics

1 answer:

serg [7]3 years ago

3 0

1.
A graph is formed by plotting all pairs (x,y), such that y is f(x), for a certain function f.

for example:

Take a look at the graph. (x, y)=(3, 5) is on the graph. This means that f(3)=5 (because f(x)=y)

2.
The real solutions of a function f, are those points x, such that f(x)=0.

in the graph we see no point (x, 0), so there are no real solutions.

Answer: <span>No Real Solution</span>

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Which equation can be used to find the perimeter of a regular octagon with sides of length 12?

USPshnik [31]

Octagans have 8 sides.
Perimeter is thought of as the length a piece of string would be if it was layed around the edges of the shape.
There are 8 sides, each 12units in length.
Perimeter=8*12=96units

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3 years ago

What is the answer to -6-(-2)=

Irina-Kira [14]

Answer:

-4

explanation:

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2 years ago

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A sphere with center A is shown. Which represents a radius?

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A because radius is half of the diameter

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2 years ago

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The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

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2 years ago

Follow this link to view Juan's work. Critique Juan's work by justifying correct solutions and by explaining any errors he made

Aleksandr [31]

Answer:

I don't no sorry please bro sorry

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2 years ago