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Zolol [24]
3 years ago
11

Which is larger 3kL or 30000L

Mathematics
1 answer:
dolphi86 [110]3 years ago
7 0
30000L is larger than 3kL
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Can somebody please help me
olasank [31]

Answer:

70

Step-by-step explanation:

Complementary angles equal 90 degrees

90 - 20 = 70

I hope this helps!

5 0
2 years ago
A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u
Kryger [21]

Given: f(x) = \frac{1}{x-2}

           g(x) = \frac{2x+1}{x}

A.)Consider

f(g(x))= f(\frac{2x+1}{x} )

f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}

f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}

f(\frac{2x+1}{x} )=\frac{x}{1}

f(\frac{2x+1}{x} )=1

Also,

g(f(x))= g(\frac{1}{x-2} )

g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}

g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}

g(\frac{1}{x-2} )= \frac{x }{1}

g(\frac{1}{x-2} )= x


Since, f(g(x))=g(f(x))=x

Therefore, both functions are inverses of each other.


B.

For the Composition function f(g(x)) = f(\frac{2x+1}{x} )=x

Since, the function f(g(x)) is not defined for x=0.

Therefore, the domain is (-\infty,0)\cup(0,\infty)


For the Composition function g(f(x)) =g(\frac{1}{x-2} )=x

Since, the function g(f(x)) is not defined for x=2.

Therefore, the domain is (-\infty,2)\cup(2,\infty)



8 0
3 years ago
Carlos has 4/5 of a tank of fuel In his car. he uses 1/10 of a tank per day. How many days will his fuel last?
damaskus [11]

Answer:

8 days

Step-by-step explanation:

4/5= 8/10

8 0
3 years ago
The product of 7 and a number is 126. Find the number
fenix001 [56]

Answer:

18

Step-by-step explanation:

Product of 7 and a number is 126.

Let's form the equation,

→ 7 × x = 126

The required value of x will be,

→ 7 × x = 126

→ x = 126/7

→ [ x = 18 ]

Hence, the number is 18.

6 0
2 years ago
The density of a substance is the mass of the substance per unit of volume. The density of the element Americium is inversely pr
Viktor [21]

Answer:

(3,4)

Step-by-step explanation:

7 0
2 years ago
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