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masha68 [24]
3 years ago
15

How much candy at $1.16 a pound should be mixed with candy worth 86 cent a pound in order to obtain a mixture of 60 pounds of ca

ndy worth a dollar a pound?
Mathematics
1 answer:
galina1969 [7]3 years ago
8 0
If you had 60 pounds of candy it would be$36.66
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skelet666 [1.2K]

Answer:

y = 5/8 x

Step-by-step explanation:

Since it goes through the origin, the y intercept is zero

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N=r (A-s) what is s
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r(A-s)=N\ \ \ \ |\text{use distributive property}\\\\rA-rs=N\ \ \ \ |-rA\\\\-rs=-rA+N\ \ \ \ |:(-r)\neq0\\\\\boxed{s=\dfrac{rA-N}{r}}

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Is it possible to divide <br> 0 by 27<br> ?
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Answer:

No. Dividing by 0 is not rational, no number works.

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The table below shows some input and output values for a function named F(n). Which of the following equations describes F(n)?
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It looks like it would be C, but I'm not sure 
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Sari is teaching her best friend how to factor trinomials with a leading coefficient greater than 1. She starts with the trinomi
Eva8 [605]

Answer:

See explanation

2x^2+5x+3=(x+1)(2x+3)

Step-by-step explanation:

1. Make sure that the trinomial is written in the correct order - the trinomial must be written in descending order from highest power to lowest power. In you case, the trinomial must be written as

2x^2+5x+3

2. Decide if the three terms have anything in common, called the greatest common factor or GCF. If so, factor out the GCF. Do not forget to include the GCF as part of your final answer.

In your case, coefficients 2, 5 and 3 do not have common factors, so go to the next step.

3. Multiply the leading coefficient and the constant, that is multiply the first and last numbers together:

2\cdot 3=6

4. List all of the factors from step 3 and decide which combination of numbers will combine to get the number next to x:

6=1\cdot 6\\ \\ \bf{6=2\cdot 3}

5. After choosing the correct pair of numbers, you must give each number a sign so that when they are combined they will equal the number next to x and also multiply to equal the number found in Step 3:

2+3=5

6. Rewrite the original problem with four terms by splitting the middle term into the two numbers chosen in step 5:

2x^2+5x+3=2x^2+2x+3x+3

7. Now that the problem is written with four terms, you can factor by grouping:

2x^2 +2x+3x+3=(2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)=(x+1)(2x+3)

6 0
3 years ago
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