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Amiraneli [1.4K]
3 years ago
13

Pleas give qulified answers

Mathematics
2 answers:
NemiM [27]3 years ago
8 0
Can I please have brainiest also I promise u the answer will be true I had the same question in my Flvs and true was right and false was wrong. I hope u get a hundred on all of the other questions. 
olchik [2.2K]3 years ago
7 0
Question 1. C
Question 2. True 

Hope this Helps!!!!
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What is the angle of Z?
gizmo_the_mogwai [7]

Answer:

Angle Z is 33.

Step-by-step explanation:

Angle Z is also 33. X and Z are equivalent angle therefore it has to be 33 as well.

3 0
3 years ago
What is 7(x + 6) solve it
Elis [28]

Answer:

49

Step-by-step explanation:

First you must do x times 6, x is really just 1. After you do that, you times it by 7.

8 0
3 years ago
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PLEASE HELP ASAP!!<br> means a lot!!
egoroff_w [7]
The correct answer is:
B) \: y + 1 =  \frac{4}{3} (x - 9)
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3 years ago
5. Which number lies between -5 and -6 on a number line? *
jeyben [28]
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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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