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myrzilka [38]
3 years ago
13

The shape of the solution region for the system (picture) can be described as which of the following? Select all that apply.

Mathematics
1 answer:
grandymaker [24]3 years ago
5 0
We have 
x>=-2
x<=6
y>=-1
y<=4

using a graph tool
see the attached figure

the solution is the figure of a rectangle <span>limited by points
</span>(-2,-1) (6,-1) (-2,4) (6,4)


the answer is 
a rectangle
a parallelogram (
rectangles<span> are subsets belonging to the set of </span><span>parallelograms)</span>

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A trout lurking 32 cm below the surface of a lake spies an insect flying 17cm above the lake. How many centimeters would the tro
Bumek [7]

Answer:

The Trout will have to jump 49 cm in order to catch the insect

Step-by-step explanation:

Here, we want to calculate the distance the Trout has to jump in order to catch the Insect

To calculate this, we need to know the difference in the distances

From what we have, the Trout is 32 cm below the surface while the Insect is 17 cm above the surface of the lake

The difference in height which will represent the distance that the Trout has to jump to catch the insect will be ;

17 + 32 = 49 cm

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2 years ago
NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
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Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

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Step-by-step explanation:

x3-10oo Dimensions ft each S.

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Just for future reference always put ^ for exponents. The answer is y-6 though. 
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