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3 years ago

Clive is making hamburgers for a cookout. How many 1/4- pounds hamburgers can he make from 2 and 3/4 pounds of ground beef

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

Answer:

11 1/4- pounds hamburgers can be made from 2 and 3/4 pounds of ground beef.

Step-by-step explanation:

Here, the amount of beef uses in 1 hamburger = 1/4 pound = 0.25 pounds

Total amount of beef available = 2 3/4 pounds

Now, solving the mixed fraction,we get

$2\frac{3}{4} = 2 + \frac{3}{4} = 2 + 0.75= 2.75$

So, the total available beef = 2.75 pounds

Now, $\textrm{ Number of burgers made} = \frac{\textrm{Weight of available beef}}{\textrm{Beef used in 1 hamburger}}$

= $\frac{2.75}{0.25} = 11$

So, <u>the total number of hamburgers made = 11 </u>

Hence, 11 1/4- pounds hamburgers can be made from 2 and 3/4 pounds of ground beef.

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In a recent census, the population of a country was 7,946,700.

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7.9467 x 10^6

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2 years ago

Can you subtract 8 tenths from 1 tenth?

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3 years ago

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Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

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Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

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$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

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$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

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for $a = 1 , c=6$ we have

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for $a = 2 , c=3$ we have

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for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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