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Snowcat [4.5K]
2 years ago
11

A local hamburger shop sold a combined total of 649 hamburgers and cheeseburgers on Friday. There were 51 fewer cheeseburgers so

ld than hamburgers. How many hamburgers were sold on Friday?
Mathematics
1 answer:
katrin2010 [14]2 years ago
7 0

Answer: 350

Step-by-step explanation:

Let the cheeseburger be represented by a.

Since there were 51 fewer cheeseburgers sold than hamburgers. This means that hamburgers will be:

= a + 51.

Since the local hamburger shop sold a combined total of 649 hamburgers and cheeseburgers. This will be:

a + a + 51 = 649

2a + 51 = 649

2a = 649 - 51

2a = 598

a = 598/2

a = 299

299 cheeseburger were sold

Therefore, the number of hamburger sold will be:

= 649 - 299

= 350

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Use the distributive property to write an expression equivalent to 8(35).
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Answer:

280

Step-by-step explanation:

All you have to do is multiply 35 by 8 to get 280.

I am joyous to assist you anytime.

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3 years ago
1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

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Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

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The function is decreasing on the interval (-∞,0) ∪ (0,1]

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3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

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The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

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The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

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