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icang [17]
3 years ago
10

What is 5/20 as in a percent

Mathematics
2 answers:
m_a_m_a [10]3 years ago
6 0
It as a percent would be 25%

Law Incorporation [45]3 years ago
5 0
\sf  \dfrac{5 \div 5}{20 \div 5} =  \dfrac{1}{4}  \\  \\ 1 \div 4  \\  \\ 0.25 \\  \\ 0.25 * 100 = 25 \%
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A parallelogram is ______ a rhombus. never, always, sometimes​
Nataly [62]

Answer:

never

Step-by-step explanation:

5 0
3 years ago
Please help!! I don’t understand this at all!!
Aleks04 [339]
B. (Not sure)

(5+A)
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(a-1)
7 0
2 years ago
The graph of the piecewise function f(x) is shown.
Shtirlitz [24]

Answer:

first one

Step-by-step explanation:

all value between 1 to 5 except 5 is shown on graph

6 0
2 years ago
Read 2 more answers
If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?
Keith_Richards [23]
\bf \begin{cases}
f(x)=2-x^{12}\\
g(x)=x^2-9\\
g(x)\div f(x)=\frac{g(x)}{f(x)}
\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\
-------------------------------\\\\
\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}

so, the domain is all real numbers EXCEPT that one.
4 0
3 years ago
Steps 2,4 and 6 are all justified for the same reason, what is the reason ?
mel-nik [20]

Because both of them are congruent triangles.

The have sides that equal to each other, angles that equal to each other and a common base.

5 0
3 years ago
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