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AURORKA [14]
2 years ago
15

During hibernation, a garter snakes body temperature never goes below 3 degrees Celsius. Write an inequality that represents thi

s situation. Plz help me with this question!!!
Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
6 0
<span>During hibernation, a garter snakes body temperature never goes below 3 degrees Celsius. Write an inequality that represents this situation.

t ≥ 3
</span>
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A town is planning a playground. It wants to fence in a rectangular space using an existing wall. What is the greatest area it c
777dan777 [17]

Answer:

1250 ft^{2}

Step-by-step explanation:

We are given that the playground is fenced on three sides of the playground and the four side has an existing wall.

Let the length of the rectangle be 'X' feet and width be 'Y' feet.

As the fencing is done using 100 feet of fence. We get the relation between the sides and the fence as, X + 2Y = 100.

As, X + 2Y = 100 → X = 100 - 2Y

Now, the area of the rectangle = XY = X × ( 100 - 2Y ).

i.e Area of the rectangle = -2Y^{2} +100Y.

The general form of a quadratic equation is y=ax^{2}+bx+c.

The maximum value of a quadratic equation is given by x=\frac{-b}{2a}.

Therefore, the greatest value of -2Y^{2} +100Y is at Y = \frac{-100}{2 \times -2} = \frac{-100}{-4} = 25.

Thus, Y = 25 and X = 100 - 2Y → X = 100 - 2 × 25 → X = 50.

Hence, the area of the rectangle is XY = 50 × 25 = 1250 ft^{2}.

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3 years ago
Undefined term which goes on forever in all directions and has no thickness
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Would this be a line?
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A quadrilateral has all sides and angles congruent, what name best describes the figure?
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The answer is a rectangle or a square.
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An object is moving at the constant rate of 32 feet per second.
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Find the volume of a cone with a height of 3.25 centimeters and<br> a diameter of 2.2 centimeters.
saveliy_v [14]

Answer:

As per Provided Information

Height of cone h is 3.25 cm

Diameter of cone is 2.2 cm

  • Radius = Diameter/2

Radius = 2.2/2

Radius = 1.1 cm

We have been asked to determine the volume of the cone .

<u>Using </u><u>Formulae </u>

\boxed{\bf \:Volume_{(Cone)}  =  \cfrac{1}{3}  \pi {r}^{2}h}

Substituting the given value and let's solve it

\sf\longrightarrow \: Volume_{(Cone)}  =  \cfrac{1}{3}  \times 3.14 \times  {1.1}^{2}  \times 3.25 \\  \\  \\ \sf\longrightarrow \: Volume_{(Cone)}  =  \cfrac{1}{3}  \times 3.14 \times 1.21 \times 3.25 \\  \\  \\ \sf\longrightarrow \: Volume_{(Cone)}  =  \cfrac{1}{3}  \times 3.14 \times 3.9325 \\  \\  \\ \sf\longrightarrow \: Volume_{(Cone)}  =  \cfrac{1}{3}  \times 12.34805 \\  \\  \\ \sf\longrightarrow \: Volume_{(Cone)}  =  \cfrac{12.34805}{3}  \\  \\  \\ \sf\longrightarrow \: Volume_{(Cone)}  = 4.11 {cm}^{3}

<u>Therefore</u><u>,</u>

  • <u>Volume </u><u>of </u><u>the </u><u>cone </u><u>is </u><u>4</u><u>.</u><u>1</u><u>1</u><u> </u><u>cm³</u><u> </u><u>.</u>
3 0
2 years ago
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