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2 years ago

13

A cafeteria starts with 750 fish sticks for lunch. Each meal they serve uses 6 of the fish sticks. At the end of the lunch there

are 42 fish sticks left. Let n be the number of meals served in the cafeteria. Set up and solve an equation to find the value of n.
I will give brainliest and i put 13 points into this

1 answer:

Firlakuza [10]2 years ago

8 0

Answer:

The value of N = 118

Step-by-step explanation:

750 = 6n + 42

move all the numbers that don't contain n to the right side

6n = 750 - 42

6n = 708

708 / 6 = 118

n = 118

plug the numbers back in

6 ( 118) = 708

708 = 708

plug the 42 back in

708 + 42 = 750

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What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where

Setler79 [48]

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$k=-16,k=-8,k=8,k=16$

Step-by-step explanation:

We are given quadratic equations as

$x^2+kx+15$

and it can be factored as

$=(x+a)(x+b)$

now, we can multiply factor term

$(x+a)(x+b)=x^2+(a+b)x+ab$

now, we can compare

$x^2+(a+b)x+ab=x^2+kx+15$

so, we get

$k=a+b$

$ab=15$

we are given that

'a' and 'b' are integers

so, we can find all possible factors

$15=(-1\times -15),(1\times 15)$

$15=(-3\times -5),(3\times 5)$

so, we can find k

At $(-1\times -15)$ :

$k=a+b$

we can plug values

$k=-1-15$

$k=-16$

At $(1\times 15)$ :

$k=a+b$

we can plug values

$k=1+15$

$k=16$

At $(-3\times -5)$ :

$k=a+b$

we can plug values

$k=-3-5$

$k=-8$

At $(3\times 5)$ :

$k=a+b$

we can plug values

$k=3+5$

$k=8$

So, values of k are

$k=-16,k=-8,k=8,k=16$

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