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3 years ago

2) Eighty students were surveyed as they left the cafeteria after lunch. The following were

Mathematics

1 answer:

KatRina [158]3 years ago

7 0

Answer:

I believe that 11 students ate only the cookie.

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Find the surface area of each figure. Round to the nearest hundredth when necessary Please quick and real answers

Vlad1618 [11]

The total surface area of the cone is 542.592 square mm

<h3>Total surface area of a cone</h3>

The formula for calculating the total surface area of a cone is expressed as:

TSA = πr(r + l)

where

r is the radius of the cone

l is the slant of the cone

Given the following parameter

l = 13.6mm

r = 8mm

Substitute

TSA = π(8)(8+13.6)
TSA = 3.14(8)(21.6)

TSA = 542.592 square mm

Hence the total surface area of the cone is 542.592 square mm

Learn more on surface area here: brainly.com/question/76387

#SPJ1

8 0

2 years ago

Order from least to greatest. 2.54, 20/ 9 , 32/ 15 , 2.62

Nina [5.8K]

32/15, 20/9, 2.54, 2.62

3 0

3 years ago

Read 2 more answers

Anlatarak cozmicekseniz cevaplamayin:)

expeople1 [14]

Wish I could help! But maybe one day I can and hopefully there’s someone on here that and assist you!

8 0

3 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

A ships triangular signal flag has a base of 8 inches and an area of 64 square inches. What is the height of the signal flag

Firlakuza [10]

A = (1/2)bh
64 = (1/2)(8)h
16 = h
The height is 16 inches

7 0

2 years ago