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Salsk061 [2.6K]
3 years ago
12

From least to greatest 2/3 -4 1/2 1/4 - 1/2 2 1/3

Mathematics
2 answers:
tamaranim1 [39]3 years ago
4 0

Answer: -4 1/2 , -1/2 , 1/4 , 2/3 , 2 1/3

Step-by-step explanation:

Yuri [45]3 years ago
3 0

For this case we have the following numbers:

\frac {2} {3} = 0.6667

-4 \frac {1} {2} = \frac {-8 + 1} {2} = \frac {-7} {2} = - 3.5

\frac {1} {4} = 0.25

- \frac {1} {2} = - 0.5\\2 \frac {1} {3} = \frac {3 * 2 + 1} {3} = \frac {7} {3} = 2.3333

If we order from least to greatest we have:

-3.5; -0.5; 0.25; 0.6667; 2.3333

Answer:

-4 \frac {1} {2}; -\frac {1} {2}; \frac {1} {4}; \frac {2} {3}; 2 \frac {1} {3}

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Use pie= 22/7 to approximate the circumference of a circle whose diameter is 28 inches.​
Nata [24]

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88 inches

Step-by-step explanation:

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2 years ago
Consider the graph of the linear equation 2x - 5y = 10.
jeyben [28]

Answer:

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Step-by-step explanation:

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3 years ago
Evaluate the expression (2x + y) - z when x= 3, y = 4, z =5​
Lapatulllka [165]

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3 years ago
Graph the function f(x) = x3 – 5x2 + 6x using the graphing calculator. What are the solutions to the related equation? Check all
Juli2301 [7.4K]

the solutions to the related equation are 0,2,3 .

<u>Step-by-step explanation:</u>

Here we have , function  f(x) = x3 – 5x2 + 6x . Graph of this function is given below  . We need to find What are the solutions to the related equation . Let's find out:

Solution of graph means the  value of x at which the value of f(x) or function is zero . We can determine this by seeing the graph as at what value of x does the graph intersect or cut x-axis !

At x = 0 .

From the graph , at x=0 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 0^3- 5(0)^2 + 6(0)

⇒ f(0) =0

At x = 2 .

From the graph , at x=2 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 2^3- 5(2)^2 + 6(2)

⇒ f(0) =0

At x=3 .

From the graph , at x=3 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 3^3- 5(3)^2 + 6(3)

⇒ f(0) =0

Therefore , the solutions to the related equation are 0,2,3 .

3 0
3 years ago
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