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miv72 [106K]
3 years ago
15

WILL MAKE BRAINLIEST PLEASE HELP WHAT IS THE DOMAIN AND RANGE

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
5 0

Answer:

(-∞, ∞) = domain

(-∞, 12] = range

Explanation:

Find the domain by finding where the function is defined. The range is the set of values that correspond with the domain.

Hope this helps! :)

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Help. I just started this​
inn [45]
X is (4/3,0)
Y is (0,-4/5)


Hope that helps!
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3 years ago
The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss a
Semenov [28]
Those wind chill formulas are REALLY complex.
Attached are the formulas for wind chill (F and C) plus the OLD wind chill formula.
Here's a wind chill calculator I wrote. http://www.1728.org/windchil.htm

(I figured you might need all the help you can get.)


5 0
4 years ago
What is the following sum?<br>(please show how you worked it out)
AleksAgata [21]

Answer:

4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )

Step-by-step explanation:

Let's start by breaking down each of the radicals:

\sqrt[3]{16x^3y}

Since we're dealing with a cube root, we'd like to pull as many perfect cubes out of the terms inside the radical as we can. We already have one obvious cube in the form of x^3, and we can break 16 into the product 8 · 2. Since 8 is a cube root -- 2³, to be specific, we can reduce it down as we simplify the expression. Here our our steps then:

\sqrt[3]{16x^3y}\\=\sqrt[3]{2\cdot8\cdot x^3\cdot y}\\=\sqrt[3]{2} \sqrt[3]{8} \sqrt[3]{x^3} \sqrt[3]{y} \\=\sqrt[3]{2} \cdot2x\cdot \sqrt[3]{y} \\=2x\sqrt[3]{2}\sqrt[3]{y}

We can apply this same technique of "extracting cubes" to the second term:

\sqrt[3]{54x^6y^5} \\=\sqrt[3]{2\cdot27\cdot (x^2)^3\cdot y^3\cdot y^2} \\=\sqrt[3]{2}\sqrt[3]{27} \sqrt[3]{(x^2)^3} \sqrt[3]{y^3} \sqrt[3]{y^2}\\=\sqrt[3]{2}\cdot 3\cdot x^2\cdot y \cdot \sqrt[3]{y^2} \\=3x^2y\sqrt[3]{2} \sqrt[3]{y}

Replacing those two expressions in the parentheses leaves us with this monster:

2(2x\sqrt[3]{2}\sqrt[3]{y})+4(3x^2y\sqrt[3]{2} \sqrt[3]{y})

What can we do with this? It seems the only sensible thing is to look for terms to factor out, so let's do that. Both terms have the following factors in common:

4, \sqrt[3]{2} , x

We can factor those out to give us a final, simplified expression:

4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )

Not that this is the same sum as we had at the beginning; we've just extracted all of the cube roots that we could in order to rewrite it in a slightly cleaner form.

6 0
3 years ago
Bonita is painting walls. Each wall is the same area and requires 2/5 gallon of paint.
shutvik [7]
You do 2/5 divided by 3/4 
4 0
3 years ago
What is the answer to 55&gt;-7x + 6
IRISSAK [1]
55>-7x + 6
Get -7x by itself
Subtract 6 from both sides
49>-7x
Divide by -7 (Remember to flip the inequality sign when dividing by negatives)
-7
5 0
3 years ago
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