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3 years ago

The graph of a quadratic function is shown. Check all of the following statements that are true of the function.

Mathematics

1 answer:

Fed [463]3 years ago

4 0

Answer:

a < 0

The vertex is (1,2)

Step-by-step explanation:

Looking at the graph

we have a vertical parabola open downward

The leading coefficient is negative

The vertex is a maximum

The vertex is the point (1,2)

The axis of symmetry is equal to the x-coordinate of the vertex, so the axis of symmetry is x=1

The y-intercept is the point (0,1)

The function has two real solutions (zeros of the function) one positive and one negative

The positive zero of the function is greater than 2

The negative zero of the function is greater than -1

<u><em>Verify each statement</em></u>

case 1) a < 0

The statement is true

Because the parabola open downward

case 2) The vertex is (1,2)

The statement is true

The vertex in this problem is the maximum of the function

case 3) The axis of symmetry is y =2

The statement is false

Because the axis of symmetry is x=1

case 4) x = 2 is a zero of the function

The statement is false

Because the positive zero of the function is greater than 2

case 5) (1,2) is a minimum

The statement is false

Because the vertex in this problem is a maximum

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Step-by-step explanation:

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3 years ago

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Answer:

$\frac{1}{ {n} + {2}}$

Step-by-step explanation:

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3 years ago

After being treated with chemotherapy, the radius of the tumor decreased by 23%. What is the corresponding percentage decrease i

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Answer:

The volume of the tumor experimented a decrease of 54.34 percent.

Step-by-step explanation:

Let suppose that tumor has an spherical geometry, whose volume ( $V$ ) is calculated by:

$V = \frac{4\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the tumor.

The percentage decrease in the volume of the tumor ( $\%V$ ) is expressed by:

$\%V = \frac{\Delta V}{V_{o}} \times 100\,\%$

Where:

$\Delta V$ - Absolute decrease in the volume of the tumor.

$V_{o}$ - Initial volume of the tumor.

The absolute decrease in the volume of the tumor is:

$\Delta V = V_{o}-V_{f}$

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The percentage decrease is finally simplified:

$\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%$

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$\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%$

$\%V = 54.34\,\%$

The volume of the tumor experimented a decrease of 54.34 percent.

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