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3 years ago

Rewrite f(x)=5/x-4+7 in fractional form

Mathematics

1 answer:

Sladkaya [172]3 years ago

7 0

rewrite f(x)=5/x-4+7 in fractional form 

Answer:

f(5/x+3)=14x+15/5+3x

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what is the probability that a randomly chosen number is less than or equal to 90 and is a multiple of 17

kodGreya [7K]

Answer:

let's see...

we have total possible outcomes.. = 90 ( from 1 to 90)

Desiree outcomes..= 5 ( 17,34,51,68,85)

and therefore.. the probability is 5/90 = 1/18.

here you go.. that should be correct I hope.

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2 years ago

Solve. Check the solution. t/8 = 15 *

sp2606 [1]

Answer:

the answer is 15 I'm positive

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3 years ago

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This figure is the net

Afina-wow [57]

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3 years ago

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Find the value of x (desperately need help with this please! Due today!)

Nadya [2.5K]

Answer:

x=48

Step-by-step explanation:

the 2 given angles if added together we know it’ll be 180 degrees

3x-12+x=180

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+12. +12

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1 year 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

4 years ago

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Rewrite f(x)=5/x-4+7 in fractional form ​

Rewrite f(x)=5/x-4+7 in fractional form