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elixir [45]
3 years ago
11

A local hamburger shop sold a combined total of 557 hamburgers and cheeseburgers on Friday. There were 57 more cheeseburgers sol

d than hamburgers. How many hamburgers were sold on Friday?
Mathematics
1 answer:
vagabundo [1.1K]3 years ago
5 0

Answer:

Equation: 557=2x+57

Let x=hamburgers sold

Solution: 250 hamburgers

Step-by-step explanation:

1. Subtract 57 from both sides of the equation

2. Divide both sides of the equation by 2

3. x=250

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3 years ago
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Triangle XYZ has vertices X(–1, –3), Y(0, 0), and Z(1, –3). Malik rotated the triangle 90° clockwise about the origin. What is t
OLga [1]

Answer:

Option D.

Step-by-step explanation:

It is given that triangle XYZ has vertices X(–1, –3), Y(0, 0), and Z(1, –3).

Malik rotated the triangle 90° clockwise about the origin. So, rule of rotation is

(x,y)\to (y,-x)

Now,

X(-1,-3)\to X'(-3,-(-1))=X'(-3,1)

Y(0,0)\to Y'(0,0)

Z(1,-3)\to Z'(-3,-(1))=Z'(-3,-1)

The image points are X’(–3, 1), Y’(0, 0), Z’(–3, –1).

Therefore, the correct option is D.

6 0
3 years ago
I have the answer for x I just need to know how to get the answer for Y, please help, thanks.
Ad libitum [116K]
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4 0
3 years ago
The population of a town grows at a rate proportional to the population present at time t. the initial population of 500 increas
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Evaluate this on your calculator.
5 0
3 years ago
A metallurgist has one alloy containing 32% copper and another containing 65% copper. How many pounds of each alloy must he use
kompoz [17]

Answer:

<h2>30.67 pounds of 32% copper alloy and 10.33 pounds of 65% copper alloy</h2>

Step-by-step explanation:

        First alloy contains 32% copper and the second alloy contains 65% alloy.

We wish to make 44 pounds of a third alloy containing 42% copper.

       Let the weight of first alloy used be x in pounds and the weight of second alloy used be y in pounds.

       Total weight = 44\text{ }pounds=x+y        -(i)

      Total weight of copper = 42\%\text{ of 44 pounds = }32\%\text{ of }x\text{ pounds + }65\%\text{ of }y\text{ pounds }

       \dfrac{42\times 44}{100}=\dfrac{32x}{100}+\dfrac{65y}{100}\\\\ 32x+65y=1848        -(ii)

       Subtracting 32 times first equation from second equation,

32x+65y-32x-32y=1848-32\times44\\33y=440\\y=13.333\text{ }pounds \\x=30.667\text{ }pounds

∴ 30.67 pounds of first alloy and 13.33 pounds of second alloy were used.

4 0
3 years ago
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