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Sonbull [250]
3 years ago
7

Need help with problem

Mathematics
1 answer:
seraphim [82]3 years ago
5 0
1. The puppy gained 3/8 pounds each week, for 4 weeks. To find out how much the puppy gained altogether, you multiply 3/8 by 4.

In order to do that, we have to turn 4 into a fraction. We can simply write it as 4/1.

Now, we multiply the two fractions together. When you multiply fractions, you multiply the numerators (top numbers) together, and the denominators (bottom numbers) together.

\frac{3}{8}* \frac{4}{1}

3 x 4 = 12

8 x 1 = 8

The puppy gained 12/8 pounds during the 4 weeks.

3. Ivana has 3/4 pounds of meat and makes 3 patties, all weighing the same.

To figure out how much each patty weighs, we first have to turn 3 into a fraction. Much like the first problem, we can simply write it as 3/1.

Then, we have to divide 3/4 by 3/1. In order to do that, we multiply 3/4 by the reciprocal of 3/1. The reciprocal is the fraction flipped upside down, so in this case, the reciprocal of 3/1 is 1/3.

Now, we multiply.

\frac{3}{4} * \frac{1}{3}

3 x 1 = 3
4 x 3 = 12

we can simplify 3/12 by dividing the numerator and denominator by 3

3 ÷ 3 = 1
12 ÷ 3 = 4

Each patty weighs 1/4 pounds
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Answer:

3

Step-by-step explanation:

When 59,527 is divided by 23, the quotient is 2,588 R3.

Hope this helps! :)

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3 years ago
Let ​g(x)x and ​h(x)<br> . Evaluate .
Maurinko [17]

9514 1404 393

Answer:

  h(g(x)) = (1/5)x² +4/5

Step-by-step explanation:

  (h\circ g)(x)=h(g(x))=h(x^2+2) = \dfrac{(x^2+2)+2}{5}=\dfrac{x^2+4}{5}\\\\\boxed{(h\circ g)(x)=\dfrac{1}{5}x^2+\dfrac{4}{5}}

3 0
3 years ago
Which ordered pair is the solution to the system of equations below using the elimination method?
kodGreya [7K]
<span>15x-3y=27
5x+4y=14

multiply (2nd) equation by (3)
15x + 12y = 42

</span>15x - 3y=27
15x + 12y = 42
-------------------substract
-15y = -15
     y = 1

15x-3y=27
15x-3(1)=27
15x - 3 = 27
15x = 30
    x = 2

so now you have x = 2, y =1

answer <span>D. (2,1)</span>


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balu736 [363]
I dont know i will tell you when i find out
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