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3 years ago

Please help and write a proper answer I really need it. Thanks.

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Answer:

A cone is 360 triangles rotated in a circle, so find the hypotenuse of the triangle and multiplying it by 360 you would get the surface area of the cone

(っ◔◡◔)っ ♥ Hope It Helps ♥

PolarNik [594]3 years ago

6 0

Answer:The value of the surface area (in square centimeters) of the cone is equal to the value of the volume (in cubic centimeters) of the cone. The formula for the surface area S of a cone is S=πr2+πrl, where r is the radius of the base and l is the slant height.

Step-by-step explanation:

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Artist 52 [7]

Price is reduced by 45%. Which means that the new price is 55% of the original.
Therefore, (55/100)*17 = $9.35 =new price

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4 years ago

Graph the line with the equation y=-x-4

Zielflug [23.3K]

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3 years ago

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4 years ago

A sample of customers from Barnsboro National Bank shows an average account balance of $315 with a standard deviation of $87.A s

cestrela7 [59]

Answer:

A) Barnsboro Bank has more variation.

Step-by-step explanation:

Coefficient of Variation:

The coefficient of variation is given by the division of the standard deviation by the mean.

In this question:

The account balance with the highest coefficient of variation has more variation.

Barnsboro National Bank shows an average account balance of $315 with a standard deviation of $87.

$cv = \frac{87}{315} = 0.2762$

Wellington Savings and Loan shows an average account balance of $8350 with a standard deviation of $1800.

$cv = \frac{1800}{8350} = 0.2156$

Barnsboro Bank has the higher cv, so more variation, and the answer is given by option A.

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3 years ago

Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y e

sergiy2304 [10]

Answer:

$y(t) = - 5 {e}^{2\pi - 2t} \cos(t)$

Step-by-step explanation:

The given initial value problem is

$y''+4y'+5=0$

$y( \frac{ \pi}{2} ) = 0$

$y'( \frac{ \pi}{2} ) = 5$

The corresponding characteristic equation is

${m}^{2} + 4m + 5 = 0$

$m = - 2 \pm \: i$

The general solution becomes:

$y(t) = A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \sin(t)$

We differentiate to get:

$y'(t) = - A {e}^{ - 2t} \sin(t) - 2 A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \cos(t) - 2B {e}^{ - 2t} \sin(t)$

We apply the initial conditions to get;

$y( \frac{\pi}{2} ) = A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

$A {e}^{ - 2\pi} (0 ) + B {e}^{ - 2\pi} ( 1) = 0$

$B = 0$

Also;

$y'( \frac{\pi}{2} ) = - A {e}^{ - 2\pi} \sin( \frac{\pi}{2} ) - 2 A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) - 2B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$