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2 years ago

15

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]2 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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If CD I EF, what is the mZFCD?

Margaret [11]

there is no picture. Can you upload a picture?

5 0

3 years ago

2. A bag contains 25 paise coins and 50 paise Coins. The number of 25 paise coins is 6 times that of so paise Coins The total va

Karo-lina-s [1.5K]

<h2>~<u>Solution</u> :-</h2>

Here, it is given that the bag contains 25 paise coins and 50 paise coins in which, 25 paise coins are 6 times than that of 50 paise coins. Also, the total money in the bag is Rs. 6.

Hence, we can see that, here, we have been given the linear equation be;

Let the number of coins of 50 paise will be $ x $ and the number of coins of 25 paise will be $ 6x $ as given. . .

Hence,

$x + 6x = 6$

$7x = 6$

$x = \frac{6}{7} \\$

$x = 1.1666667 \approx \: 2$

Hence, the number of 50 paise coins will be <u>2</u>. And, 6 times of two be;

$6(2)$

Hence, the number of 25 paise coins will be <u>12</u>.

6 0

2 years ago

Simplify 5 square root of 7 end root plus 12 square root of 6 end root minus 10 square root of 7 end root minus 5 square root of

Sphinxa [80]

The simplest answer form would be -5 square root of 7 end root + 7 square roots of 6

4 0

3 years ago

g Given the system of equations: 15 c1 – 5 c2 – c3 = 2400 - 5 c1 + 18 c2 – 6 c3 = 3500 - 4 c1 - c2 + 12 c3 = 4000 (a) Calculate

denpristay [2]

Answer

(a)

$A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}$

(b)

$A^{-1} b = \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}$

Step-by-step explanation:

Remember that when you want to solve a problem like this, you express the equation as following

$Ax = b$

So, if you know the inverse of $A$ then

$x = A^{-1} b$

For this case

$A = \begin{pmatrix} 15 && 5 && -1 \\ -5 && 18 && -6 \\ -4 && -1 && 12 \end{pmatrix}$

Now for this case the inverse of A would be

$A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}$

Then when you multiply with the vector solution

$A^{-1} b = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix} * \begin{pmatrix} 2400 \\ 3500 \\ 4000 \end{pmatrix} = \frac{1}{3493} \begin{pmatrix} 249500 \\ 1197600 \\ 1347300 \end{pmatrix}\\\\\\= \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}$

So from that information you can conclude that the solution to the system of equations is x = 500/7 y = 2400/7 and z = 2700/7

7 0

3 years ago

A square is inscribed in a circle of diameter 12 millimeters. What is the area of the shaded region? A square is inscribed in a

bezimeni [28]

Answer: A. (72π - 144) mm²

<u>Step-by-step explanation:</u>

$A_{shaded}=A_{circle}-A_{square}\\\\\\A_{circle}=\pi \cdot r^2\\.\qquad \ =\pi \bigg(\dfrac{12\sqrt2}{2}\bigg)^2\\\\.\qquad \ =\pi (6\sqrt2)^2\\.\qquad \ =72\pi\\\\\\A_{square}=side^2\\.\qquad \quad =\dfrac{12\sqrt2}{\sqrt2}^2\\\\.\qquad \quad =12^2\\\\.\qquad \quad =144\\\\\\\large\boxed{A_{shaded}=72\pi-144}$

7 0

3 years ago