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Natali [406]
2 years ago
12

For f(x) = 2x-1 and g(x)= x+3, find f(g(x))

Mathematics
1 answer:
Over [174]2 years ago
7 0

Answer:

C

Step-by-step explanation:

Plug in g(x) into the variable x of f(x), we get:

(fog) = 2(x+3)-1 = 2x+6-1=2x+5

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CS Algebra
CaHeK987 [17]

Answer:

Hope this helps :)

Step-by-step explanation:

8(x - 2) = 2x + 8

y+9 = -2(y + 1)

value of x in 8(x - 2) = 2x + 8

x=4

substitue

y+9=−2(y+1)

value of y y+9=−2(y+1)

y= - 11/3 or 3.66...

x=4

y=4 (I rounded 3.66)

4 0
2 years ago
Plz solve i need this done today
garik1379 [7]
D,A,B,C i think this is correct !
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What is the missing angles
Vaselesa [24]
K = 42, m = 46, j =92
6 0
3 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
WILL
IRISSAK [1]

Answer:

-10, -|-8|, -(-2), |-11|, 5²

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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