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Serhud [2]
3 years ago
8

give an example if an equation where any value if X will make the equation true. Infinite number of solutions. PLS HELP

Mathematics
1 answer:
defon3 years ago
8 0

Answer:

x+3 = x+3

Step-by-step explanation:

No matter what you plug into x, both sides will be the same.

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The ordered pair of y=2x+1 & y=-x+4
8_murik_8 [283]
I hope this helps you


-x+4=2x+1


4-1=2x+x


3=3x


x=1


y=-1+4


y=3
8 0
3 years ago
Classify the triangle based on its sides.<br><br> A equilateral C iscoles<br><br> B scalene
dangina [55]

Answer:

Step-by-step explanation:

The answer is scalene triangle

8 0
3 years ago
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Solve. -1/2 + (3/4 x 4/9)
max2010maxim [7]
-0.16666666666 which equals -1/16.
4 0
3 years ago
What is the value of the expression 9+4n-1 when n = 3?
dangina [55]
Answer: 9+4n-1 = 20


We can solve this by substitution.
Replace n with the value given, 3 (remember 4n means 4 times n):
9 + 4*3 - 1
Then work it out using arithmetic
9+12-1
=20
6 0
2 years ago
Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2 ( using the property (ab)^m=a^m.b^m )

=(2^6)(3^{-12})\times (2^{-6})(3^4) ( using the property (a^m)^n=a^{mn}

=(2^6)(2^{-6})(3^{-12})(3^4) ( combining the like powers )

=2^{6-6}3^{-12+4} ( using the property a^m.a^n=a^{m+n} )

=2^03^{-8}

=\frac{1}{3^8} ( using the property a^{-m}=\frac{1}{a^m} )

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

6 0
3 years ago
Read 2 more answers
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