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3 years ago

A package weighs 96 ounces what's the weight of the packet in pounds?

Mathematics

2 answers:

Sauron [17]3 years ago

8 0

I think the weight of the packet in pounds is 6 pound i]m not sure

Nezavi [6.7K]3 years ago

5 0

16 ounces is equal to a pound. To find how many pounds, you need to divide 96 by 16. 96 divided by 16 is 6. 96 ounces is equal to 6 pounds.

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Answer:

A requirement necessary for a given statement or theorem to hold. Also called a condition.

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3 years ago

1/4 (g + 2) = 8<br><br> Thank you

Irina-Kira [14]

Answer:

Step-by-step explanation:

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3 years ago

Sara bought 4 pounds

Roman55 [17]

Answer:

$12.80

Step-by-step explanation:

Sara bought 4 pounds of coffee.

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4 years ago

Read 2 more answers

Help i dont understand how to complete this

Luda [366]

Answer: Josh = 19.8 hours, Danny = 10.8 hours

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

$Josh: \dfrac{1}{x+9}\\\\\\Danny: \dfrac{1}{x}\\\\\\Together: \dfrac{1}{7}\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\\dfrac{1}{x+9}\quad +\qquad \dfrac{1}{x}\qquad = \qquad \dfrac{1}{7}\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(1)(-63)}}{2(1)}\\\\\\.\quad =\dfrac{5\pm\sqrt{25+252}}{2}\\\\\\.\quad =\dfrac{5\pm\sqrt{277}}{2}\\\\\\.\quad =\dfrac{5\pm 16.6}{2}\\\\\\x =\dfrac{5+16.6}{2}\qquad x=\dfrac{5-16.6}{2}\\\\\\x=\dfrac{21.6}{2}\qquad \qquad x=\dfrac{-11.6}{2}\\\\\\x=10.8 \qquad \qquad x=-5.8$

Since time cannot be negative, x=-5.8 is an extraneous solution (not valid) so x = 10.8

Josh: x + 9 --> 10.8 + 9 = 19.8

Danny: x --> 10.8

****************************************************************************************

2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3 <em>(x = -1 is an extraneous solution so is not valid)</em>

2d) No Solution <em>(x = 1 is an extraneous solution so is not valid)</em>

Here is the work for 2a. Follow this format for b, c, & d

$\dfrac{3}{k^2-8x+12}=\dfrac{k}{k-2}-\dfrac{4}{k-6}\\\\\text{Since the denominator cannot equal zero, then } k \neq2\ and\ k\neq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}$

$x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(5)}}{2(1)}\\\\\\.\quad =\dfrac{10\pm\sqrt{100-20}}{2}\\\\\\.\quad =\dfrac{10\pm\sqrt{80}}{2}\\\\\\.\quad =\dfrac{10\pm8.9}{2}\\\\\\x=\dfrac{10+8.9}{2}\qquad x=\dfrac{10-8.9}{2}\\\\\\x=\dfrac{18.9}{2}\qquad x=\dfrac{1.1}{2}\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}$

8 0

3 years ago

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Dahasolnce [82]

Answer:

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Step-by-step explanation:

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