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White raven [17]
1 year ago
8

Determine the Roman and range for the relation in interval notation

Mathematics
1 answer:
Ivan1 year ago
5 0

Domain:

(-\infty,\infty)

Range:

(-\infty,\infty)

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What is the the greatest common factor of 10m" and 4m”.
Ilya [14]

Answer:

2

Step-by-step explanation:

4 dvided by 2 is 2

10 didide by 2 is 5

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3 years ago
There were 80 tickets sold for the class play. Of these,24 adults and the rest were children . Write the ratio if adults to chil
nevsk [136]

Answer:

ratio of adults to children in SIMPLEST form would be 3:7

Step-by-step explanation:

To simplify a ratio you have to find the Greatest Common Factor (GCF) of each number in the ratio and divide. In this example, the original ratio would be 24:56 the GCF of 24 would be 9, and the GCF of 56 would be 8. And then you would divide those numbers and get 3:7

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3 years ago
Help me please please I’ll give brainly
Mariana [72]

Answer:

the correct answer is B

Step-by-step explanation:

4 0
2 years ago
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The base radius of an ice cream cone is 3.5 cm and the slant height is 6.5 cm. What is the capacity of the ice cream cone and it
ella [17]
The formula of cone's capacity is V = (1 / 3) (pi x r x h)
we know that r = 3.5, we must find h, the slant height is H=6.5, applying pythagorean theorem  h ^2 = H^2, so h= sqrt (H^2 -  r^2)
h =  sqrt (6.5^2 - 3.5^2) = 5.47
 finally, V = (1 / 3)  (3.14 x 3.5 x 5.47) = 20.03 cm^3
 the surface are is A = pi x r x H = 71.43 cm^2
8 0
3 years ago
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Which choice is equivalent to the expression below when y2 0?<br> √y^2 + √16y^3 – 4y√y
DanielleElmas [232]

Answer:

Option C.

Step-by-step explanation:

We start with the expression:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

where y > 0. (this allow us to have y inside a square root, so we don't mess with complex numbers)

We want to find the equivalent expression to this one.

Here, we can do the next two simplifications:

\sqrt{16*y^3} = \sqrt{16} \sqrt{y^3} = 4*\sqrt{y^3}

And:

y*\sqrt{y} = \sqrt{y^2} *\sqrt{y} = \sqrt{y^2*y} = \sqrt{y^3}

If we apply these two to our initial expression, we can rewrite it as:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

\sqrt{y^3}  + 4*\sqrt{y^3} - 4\sqrt{y^3} = \sqrt{y^3}

Here we can use the second simplification again, to rewrite:

\sqrt{y^3} = y*\sqrt{y}

So, concluding, we have:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y} = y*\sqrt{y}

Then the correct option is C.

8 0
3 years ago
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