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Veseljchak [2.6K]
3 years ago
11

Which is bigger 96oz or 5 lbs

Mathematics
2 answers:
Mariana [72]3 years ago
5 0

9​6 ounces is equal to 6 pounds therefore bigger than 5 pounds

lapo4ka [179]3 years ago
3 0

There are 16 oz in 1 pound.

Multiply 5 lbs with 16 oz

5 x 16 = 80

96 > 80, therefore, 96 oz is greater than 5 lbs

96 oz is your answer

hope this helps

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Li Ping says a square with 3 inch sides has greater area than a parallelogram that is not a square but has sides that have the s
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Li Ping's statement makes sense.

Step-by-step explanation:

The area of a square with side lengths a inches is given by, A_1 = a^{2}.

Now, the area of a parallelogram with equal sides a inches, which is not a square is given by, A_2 = ah, where, h is the perpendicular distance between the opposite sides.

See the diagram attached.

Since a is the hypotenuse of as right triangle with height h, hence, a > h.

So, A_1 > A_2

Therefore, Li Ping's statement makes sense. (Answer)

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3 years ago
What is the mean absolute deviation of astroids data to answer the question first find the mean next fill in the table for the a
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Step-by-step explanation:

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3 years ago
= 2. Given the quadratic equation 3x2 – 8x + k = 0 has no real roots. Find the range of values of k.
VashaNatasha [74]

\text{If there  are no real roots, then}\\\\~~~~~\text{Discriminant} < 0\\\\\implies b^2 -4ac < 0\\\\\implies (-8)^2 -4 \cdot 3 k < 0\\\\\implies 64-12k < 0\\\\\ \implies 12k > 64\\\\\implies k > \dfrac{64}{12}\\\\\implies k > \dfrac{16}3\\\\\text{Interval,}~ \left( \dfrac{16}3, ~\infty\right)

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East Middle's football team was positioned on the 50 yard line. After a penalty is called, the football team is pushed back 5 ya
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Answer:

The answer would be the 35 yard line.

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Read 2 more answers
Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0
3 years ago
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