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3 years ago

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Campbell's soup is creating a new soup label. if a can has a height of 6 in and a diameter of 4 in, how much material does campb

ell need for each soup label ?

1 answer:

nasty-shy [4]3 years ago

8 0

Assuming the soup label covers the entire surface of the can, we solve for surface area. S.A. = 2pi*r^2 + pi*d*h = 8pi + 24pi = 32pi

32pi

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Which expression has a value of 7/12 ? 7/8 + 4 , 1/3 + 1/4 , 6/8 + 1/4 , 2 1/12 + 5 1/12

vazorg [7]

Answer:

$\frac{1}{3} + \frac{1}{4}$

Step-by-step explanation:

We want to find an expression that has a value of

$\frac{7}{12}$

First option:

$\frac{7}{8} + 4 = \frac{7}{8} + \frac{32}{8} = \frac{39}{8} \ne \frac{7}{12}$

Second option:

$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$

Correct option

Third option

$\frac{6}{8} + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} = 1$

Fourth option:

$2 \frac{1}{12} + 5 \frac{1}{12} = \frac{43}{6}$

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3 years ago

At a local high school, the probability that a student has a job and a car is 0.18. Jackson wants to find the probability that a

makvit [3.9K]

<span>D. the probability that a student has a car or job

</span>

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3 years ago

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2.789 in word form and in expanded form

mylen [45]

Word form:
Two and seven hundred eighty-nine thousandths

Expanded form:

2

+ 0.7

+ 0.08

+ 0.009

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7x^2 +2× when x = 3?

borishaifa [10]

Answer:

7(3^2) + 2(3).

That equals to 7(9) + 2(3).

That equals to 63 + 6.

That finally equals to 69.

Step-by-step explanation:

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A cone and a triangular pyramid have a height of 9.3 m

Otrada [13]

Answer:

$x=17.1\ in$

Step-by-step explanation:

<u><em>The complete question is</em></u>

A cone and a triangular pyramid have a height of 9.3 m and their cross-sectional areas are equal at every level parallel to their respective bases. The radius of the base of the cone is 3 in and the other leg (not x) of the triangle base of the triangular pyramid is 3.3 in

What is the height, x, of the triangle base of the pyramid? Round to the nearest tenth

The picture of the question in the attached figure

we know that

If their cross-sectional areas are equal at every level parallel to their respective bases and the height is the same, then their volumes are equal

Equate the volume of the cone and the volume of the triangular pyramid

$\frac{1}{3}\pi r^{2}H=\frac{1}{3}[\frac{1}{2}(b)(h)H]$

simplify

$\pi r^{2}=\frac{1}{2}(b)(h)$

we have

$r=3\ in\\b=3.3\ in\\h=x\ in\\pi=3.14$

substitute the given values

$(3.14)(3)^{2}=\frac{1}{2}(3.3)(x)$

solve for x

$28.26=\frac{1}{2}(3.3)(x)$

$x=28.26(2)/3.3\\x=17.1\ in$

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