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2 years ago

8

Which description best fits the graph? 33 у 3 always decreasing 2 1 increasing, then decreasing -3 -2 -1 0 1 2 ЗХ -1 -2 always i

ncreasing -3 decreasing, then increasing

1 answer:

kozerog [31]2 years ago

7 0

Answer:

Always increasing!

Step-by-step explanation:

I hope this helps better understand it! Your graph never decreases, always increasing just slows down in the middle

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Which represents the area of the shaded region?

Dimas [21]

The area of the shaded region is Area of the Circle - Area of the Triangle + Area of the Square.

Hence, option D) is the correct answer.

<h3>What is the representation of the area of the shaded region?</h3>

From the diagram below;

A triangle is inscribed in a circle
A Square is inscribe in a triangle.

Only the the circle and square are shaded.

To get the shaded region, we say;

Area of the Circle - Area of the Triangle + Area of the Square.

The area of the shaded region is Area of the Circle - Area of the Triangle + Area of the Square.

Hence, option D) is the correct answer.

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2 years ago

The quotient of 14 and 7

stepladder [879]

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3 years ago

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The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 10

serious [3.7K]

Answer:

The rate of change of the height of the box at which is decreasing is $\frac{5000}{130321}$ centimeters per second.

Step-by-step explanation:

From Geometry the volume of a rectangular box ( $V$ ), measured in cubic centimeters, with a square base is modelled by the following formula:

$V = A_{b}\cdot h$ (Eq. 1)

Where:

$A_{b}$ - Area of the base, measured in square centimeters.

$h$ - Height of the box, measured in centimeters.

The height of the box is cleared within the formula:

$h = \frac{V}{A_{b}}$

If we know that $V = 500\,cm^{3}$ and $A_{b} = 361\,cm^{2}$ , then the current height of the box is:

$h = \frac{500\,cm^{3}}{361\,cm^{2}}$

$h = \frac{500}{361}\,cm$

The rate of change of volume in time ( $\frac{dV}{dt}$ ), measured in cubic centimeters per second, is derived from (Eq. 1):

$\frac{dV}{dt} = \frac{dA_{b}}{dt}\cdot h + A_{b}\cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA_{b}}{dt}$ - Rate of change of the area of the base in time, measured in square centimeters per second.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per second.

If we get that $\frac{dV}{dt} = 0\,\frac{cm^{3}}{s}$ , $\frac{dA_{s}}{dt} = 10\,\frac{cm^{2}}{s}$ , $h = \frac{500}{361}\,cm$ and $A_{b} = 361\,cm^{2}$ , then the equation above is reduced into this form:

$0\,\frac{cm^{3}}{s} = \left(10\,\frac{cm^{2}}{s} \right)\cdot \left(\frac{500}{361}\,cm \right)+(361\,cm^{2})\cdot \frac{dh}{dt}$

Then, the rate of change of the height of the box at which is decreasing is:

$\frac{dh}{dt} = -\frac{5000}{130321}\,\frac{cm}{s}$

The rate of change of the height of the box at which is decreasing is $\frac{5000}{130321}$ centimeters per second.

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3 years ago

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Natalija [7]

X(x+2) = 48
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x^2 + 2x - 48 = 0
*Use quadratic formulae*
x = 6
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4 years ago

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kompoz [17]

Answer:

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Step-by-step explanation:

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