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2 years ago

HELPPPPPPPPPPPPP ME PLEASEEEEEE

Mathematics

1 answer:

julia-pushkina [17]2 years ago

6 0

Answer:

Step-by-step explanation:

x/7 <2

Multiply both sides if the inequality by 7

x/7×7 < 2×7

Then cancel out the 7s from x/7×7 to get x by itself

now

x < 14

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If you go on this question at all you are a republican and that is good

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2 years ago

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Elijah’s workout at the gym consist of the elliptical and treadmill. After burning 450 cal on the elliptical machine, Elijah swi

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2 years ago

Pls help

KATRIN_1 [288]

Answer:

169.04 in² (nearest hundredth)

Step-by-step explanation:

Surface area of a cone = $\pi$ r² + $\pi$ r $l$

(where r = radius of the base and $l$ = slant height)

Given slant height $l$ = 10 and surface area = 188.5

Surface area = $\pi$ r² + $\pi$ r $l$

188.5 = $\pi$ r² + 10 $\pi$ r

$\pi$ r² + 10 $\pi$ r - 188.5 = 0

r = $\frac{-10\pi +\sqrt{(10\pi )^2-(4\times\pi \times-188.5)} }{2\pi }$ = 4.219621117...

Volume of a cone = (1/3) $\pi$ r²h

(where r = radius of the base and h = height)

We need to find an expression for h in terms of $l$ using Pythagoras' Theorem a² + b² = c², where a = radius, b = height and c = slant height

r² + h² = $l$ ²

h² = $l$ ² - r²

h = √( $l$ ² - r²)

Therefore, substituting found expression for h:

volume of a cone = (1/3) $\pi$ r²√( $l$ ² - r²)

Given slant height $l$ = 10 and r = 4.219621117...

volume = 169.0431969... = 169.04 in² (nearest hundredth)

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2 years ago

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cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0

Dafna11 [192]

Answer:

Radius=2.09 cm

Height,h=14.57 cm

Step-by-step explanation:

We are given that

Volume of cylinderical shaped can=200 cubic cm.

Cost of sides of can=0.02 cents per square cm

Cost of top and bottom of the can =0.07 cents per square cm

Curved surface area of cylinder= $2\pi rh$

Area of circular base=Area of circular top= $\pi r^2$

Total cost,C(r)= $0.02\times 2\pi rh+2\pi r^2\times 0.07$

Volume of cylinder, $V=\pi r^2 h$

$200=\pi r^2 h$

$h=\frac{200}{\pi r^2}$

Substitute the value of h

$C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07$

$C(r)=\frac{8}{r}+0.14\pi r^2$

Differentiate w.r.t r

$C'(r)=-\frac{8}{r^2}+0.28\pi r$

$C'(r)=0$

$-\frac{8}{r^2}+0.28\pi r=0$

$0.28\pi r=\frac{8}{r^2}$

$r^3=\frac{8}{0.28\pi}=9.095$

$r=(9.095)^{\frac{1}{3}}=2.09$

Again, differentiate w.r.t r

$C''(r)=\frac{16}{r^3}+0.28\pi$

Substitute the value of r

$C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0$

Therefore,the product cost is minimum at r=2.09

h= $\frac{200}{\pi (2.09)^2}=14.57$

Radius of can,r=2.09 cm

Height of cone,h=14.57 cm

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3 years ago

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Answer:

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4- 105%

Step-by-step explanation:

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