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lora16 [44]
2 years ago
10

For each pair of functions F and G below, find f(g(x)) and g(f(x)).

Mathematics
1 answer:
8090 [49]2 years ago
8 0

Answer:

FG

Step-by-step explanation:

COOL AIDS BRO FAX AND MORE STUFF

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What is the ordered pair of y=-5/9x-6
antoniya [11.8K]

Answer:

B), x=-1, y=-2, z=1.

Step-by-step explanation:

7 0
3 years ago
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
How many acute angles does an obtuse triangle have​
TiliK225 [7]

Answer:

i think 2

Step-by-step explanation:

ya i think 2

6 0
2 years ago
A blank CD can hold 70 minutes of music. So far you have burned 25 minutes of music onto the CD. You estimate that each song las
zaharov [31]

Answer:

The possible numbers of songs that you can burn onto the CD is 11.

Step-by-step explanation:

The total storage capacity of CD =  70 minutes

Storage already used = 25 minutes

Hence, The available storage =  Total available  storage -  Used Storage

= 70 minutes - 25  minutes = 45 minutes

Hence, the CD has 45 minutes storage left.

Now, each song takes up 4 minutes of storage.

⇒ \textrm{Total number of songs possible} =  \frac{\textrm{Total storage left}}{\textrm{Time taken by each song}}

= \frac{45 minutes}{4minutes }  = 11.25

⇒The number of songs possible = 11.25 ≈  11

Hence, the possible numbers of songs that you can burn onto the CD is 11.

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3 years ago
The paragraph below comes from the rental agreement Susan signed when she opened her account at Super Video.
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0283646.8 dhbdvfbe 29376.0
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