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2 years ago

Weight Price 3 Oz $10.00 OZ $15.00

Mathematics

1 answer:

alukav5142 [94]2 years ago

4 0

Answer:weight and the Oz would be the same for both problems

Step-by-step explanation:

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A customer(s) needs 60 pencils. If they buy them on sale, how many of the 60 pencils can they get for free? (what the picture sa

Ray Of Light [21]

Answer: D) 12

Step-by-step explanation:

If the customer buys 4 pencils he will get the 5th one free.

This means, in every 5 pencils, he is getting one free pencil. 60 pencils can be split into 12 groups of 5 pencils each. In each group he is getting 1 free pencil, so in 12 groups, the customer will get 12 free pencils.

So, we can say that if the customer buys 60 pencils on sale, he will get 12 of them free.

5 0

2 years ago

What are the greatest common factors of 15 and 6

laiz [17]

The greatest common factor is 3.

7 0

3 years ago

Read 2 more answers

X + 3y =- 12<br> y+ 3 =- {(x – 3)

fenix001 [56]

Answer:

x = 6, y = -6

Step-by-step explanation:

solve for x and y

x + 3y = -12

y + 3 = -x + 3

add

(x + 3y) + -x + 3 = -12 + y + 3

3y + 3 = y - 9

2y + 3 = -9

2y = -12

y = - 6

x + 3(-6) = -12

x - 18 = -12

x = 6

8 0

3 years ago

Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

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3 years ago

Given the argument: Q / M v-Q // M M .Q This argument is:<br> Valid or Invalid

Sliva [168]

The answer is invalid

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3 years ago