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Anettt [7]
3 years ago
14

Get ready for the launch of America’s Top Equation! Imagine that a TV network held a talent contest for . . . equations. Which t

ype of equation would you want to win? Pick the kind of function that you’d like to sponsor for the contest, and describe its traits in a way that shows why it should be America’s top equation. Recall some of the types of equations: linear, exponential, and quadratic (you’ve seen these equations applied to increasingly advanced scenarios), plus polynomial, radical, and rational. You’ve seen these equations in systems and by themselves, and sometimes as the basis of an inequality or a system of inequalities. In your description, include key features (as represented in a graph, equation, or table) and prove your equation type’s worth for solving real-world problems. America’s top equation must prove useful and applicable to everyday situations! No Response(s)
Mathematics
1 answer:
netineya [11]3 years ago
7 0

Answer

-1(2/3*8)+23*(-19)

This is my favorite equation.


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3 years ago
Un cono ha l'area laterale di 255 pigreco cm^2, l'apotema di 17 cm e pesa 900 pigreco g. Calcola il peso specifico del materiale
valkas [14]

Answer:

The specific weight is 1.5\frac{g}{cm^{3}}

Step-by-step explanation:

The question in English

A cone has a lateral area of 255 pi cm^2, an apothem of 17 cm and weighs 900 pi g. It calculates the specific weight of the material of which it is composed

step 1

Find the radius of the cone

we know that

The lateral area of a cone is equal to

LA=\pi rl

we have

LA=255\pi\ cm^{2}

l=17\ cm

substitute the values

255\pi=\pi r(17)

Simplify

255=r(17)

r=255/(17)=15\ cm

step 2

Find the height of the cone

Applying the Pythagoras Theorem

l^{2} =r^{2} +h^{2}

substitute the values and solve for h

17^{2} =15^{2} +h^{2}

h^{2}=17^{2}-15^{2}

h^{2}=64

h=8\ cm

step 3

Find the volume of the cone

The volume of the cone is equal to

V=\frac{1}{3}\pi r^{2}h

substitute the values

V=\frac{1}{3}\pi (15)^{2}(8)

V=600\pi\ cm^{3}

step 4

Find the specific weight

Divide the mass by the volume

\frac{900\pi }{600\pi}=1.5\frac{g}{cm^{3}}

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