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2 years ago

Help with 1,2,3,4 image attached! Please explain ❤️ I will give thanks

Mathematics

1 answer:

Grace [21]2 years ago

6 0

1. x/y
4/13
5/15
2. (I'm not sure but here is my guess)
x/y
4/21
5/20
3. x/y
10/29
33/98
4. x/y
200/105
295/200
I'm pretty sure these answers are correct (at least I hope because I spent a while on these.) I hope this helps!

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How do we refer to self-contained state actions: monadic. dyadic. triadic. k-adic?

Assoli18 [71]

This state action is referred to as monadic. This is a function or a relation with an arity of one. A monad can relate an algebraic theory into a <span>composition of a function though its power is not always apparent.</span>

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3 years ago

An egg of a certain kind of insect called a Tachinidae is 0.00002 meters long. What is this length in scientific notation?

GaryK [48]

2.0x10^-5 or 2.0E-5 —this is because as a small decimal number, you move the decimal to the right five spaces and then add a negative to the decimal to indicate that when you translate back (from scientific notation to 0.00002) you will move the decimal to the left

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3 years ago

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True or False

BigorU [14]

Answer:

1. false

2. true

3. true

4. true

5. true

6. false

Step-by-step explanation:

sub to tapl :L

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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

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3 years ago

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Evaluate j/k -0.2k when j =25 and k = 5

denis-greek [22]

When you have an equation that has variables and you know the value of those variables, just plug it it. So plugging the vaules we get
$\frac{25}{5}-\frac{1}{5}5$
Just evaluating we get 4

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2 years ago

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