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4 years ago

9

The radius of a DR.PEPPER can is 3 cm and the height is 12 cm. What is the distance from the center of the bottom of the can to

the edge of the top of the can?

1 answer:

Gemiola [76]4 years ago

6 0

Isn't it just 12 cm? Wait no I think it's 11.62

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Solve.<br><br> 23x+56=313<br><br> Enter your answer as a mixed number in simplest form in the box.

dangina [55]

23x + 56 = 313

subtract both sides by 52

23x = 257

then divide both sides by 23

x = 11.1739

(approximate)

hope this helped!

8 0

3 years ago

Read 2 more answers

A cylindrical container holding sugar has a height of 15 inches and a diameter of 4 inches. Which of the following is the closes

nadya68 [22]

V=(h)(pi)(r^2)
say pi=3.14
h=15
d/2=r=4/2=2

so
v=(15)(3.14)(2)^2
v=(15)(3.14)(4)
v=(60)(3.14)
v=188.4

so answer is D

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3 years ago

Evaluate 5 - 4b if b = -3

victus00 [196]

Answer:

17

Step-by-step explanation:

5-4b ; b=-3

- substitute b= -3 into the expression

5-4×(-3)

- multiply the numbers

(-4×(-3)=+12) 5+12

- add the numbers

(5+12=17)

SOLUTION: 17

4 0

3 years ago

Consider the equation:

Ludmilka [50]

Answer:

x^2 - 6x - 55 = 0

using calculator apply mode 5 3 and write nmbrs

1 , -6 , -55

answers must be 11 and -5

4 0

3 years ago

(points on a circle) PLEASE HELP THIS IS DUE TONIGHT

VashaNatasha [74]

Answer:

Inside the circle, a distance of $\sqrt{34}$ from the center of the circle.

Step-by-step explanation:

Center of circle is at (0,0)

One point of the circle is at (0, $\sqrt{38}$ )

Since the only known point is in the y-axis, the radius of the circle must be $\sqrt{38}$ .

$\sqrt{38}$ = 6.164

So the point (-5,-3) is still inside the circle as the distance from point (-5,-3) to the center of the circle is $\sqrt{(5)^2+(-3)^2}$ or $\sqrt{34}$ which is less than the radius of the circle.

7 0

3 years ago