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ss7ja [257]
3 years ago
14

Help please how do I do this

Mathematics
2 answers:
erik [133]3 years ago
5 0
“Reflection” because it’s the same but reflected
salantis [7]3 years ago
4 0
The transaction that is being used is a reflection
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Please help ASAP is this correct?
emmasim [6.3K]

Answer:

yess that is correct

Step-by-step explanation:

18<27

7 0
3 years ago
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How do I solve this problem <br>8+1/5=
Romashka-Z-Leto [24]
Find the LCD and then combine


Answer: 41/5 or 8.2 in decimal form
5 0
3 years ago
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Write a quadratic equation with a integer coefficients with roots x= 1/2 and x=-4
Misha Larkins [42]

The quadratic equation with roots a and b is given by:

x^2-(a+b)x+ab=0

Here a=1/2 and b=-4.

So the equation is given by:

\begin{gathered} x^2-(\frac{1}{2}-4)x+(\frac{1}{2})(-4)=0 \\ x^2+\frac{7}{2}x-2=0 \\ 2x^2+7x-4=0 \end{gathered}

The quadratic equation is as shown above.

8 0
1 year ago
A gardener wants to fertilize 900 African violets. Each container of fertilizer will supply up to 65 plants. How many containers
meriva

Answer:

900÷65 = 13.86

Step-by-step explanation:

you'll put the answer how is needed

7 0
2 years ago
What is the simplified form of the expression? 6/√5−√2
tester [92]

Answer:

The expression can be simplified into 2\,\sqrt{5} +2\sqrt{2}

with no square roots in the denominator

Step-by-step explanation:

We are considering the expression  \frac{6}{\sqrt{5}-\sqrt{2}  }  which I believe is what you wanted to type in your question.

The idea behind simplification of the expression, is normally associated with the rationalization of the denominator by eliminating irrational numbers from it. This means to get rid of any "irrational" number expression (like in this example, the square root of non-perfect square numbers (\sqrt{5}\,\,\,and \,\,\sqrt{2}).

We proceed by writing this quotient in an equivalent form by multiplying its numerator and its denominator by what is called the "conjugate" of the denominator. In our case, multiply by " (\sqrt{5} +\sqrt{2}) ", because such product will allow the denominator to become a rational number:

\frac{6}{\sqrt{5}-\sqrt{2}  }\,*\,\frac{(\sqrt{5}+\sqrt{2} )}{(\sqrt{5}+\sqrt{2}) } =\frac{6\,(\sqrt{5}\,+\,\sqrt{2} )}{\sqrt{5}^2-\sqrt{2}^2 }=\frac{6\,(\sqrt{5}\,+\,\sqrt{2} )}{3}=2\,\sqrt{5} +2\sqrt{2}

where we find all square roots eliminated from the denominator

5 0
3 years ago
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