(-3/8) / (-1/3)
= (<span>-3/8) x (-3)
= 9/8
= 1 1/8</span>
(fg)(x)= x³ - 7x² - 9x + 8
domain is (-∞,∞)
Given: f(x)= x-8, g(x)= x² +1x - 1
To find (fg)(x) we multiply f(x) and g(x)
(fg)(x) = f(x) * g(x)
(fg)(x)= (x-8) (x²+x-1)
(fg)(x)= x³ + x² - x - 8x² - 8x +8
(fg)(x)= x³ - 7x² - 9x + 8
we know, domain for all cubic function is set of all real numbers
domain is (-∞,∞)
Know more about "functions" here: brainly.com/question/12431044
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9514 1404 393
Answer:
In step 4, Jim's answer is incorrect.
Step-by-step explanation:
In step 1, Jim swaps the order of addends using the commutative property of addition.
In step 2, Jim uses the distributive property to factor -1 from the final two terms. (The associative property lets Jim move parentheses.)
6.1 +(-8.5 -1.3) . . . associative property
6.1 +(-1)(8.5 +1.3) . . . distributive property
In step 3, Jim has used the properties of real numbers to form the sum of two of them.
In step 4, Jim wrote an answer of 1.1, when the answer should have been -3.7. Jim's answer is incorrect.
__
The descriptive statements about steps 2 and 4 are both true.
The answer is c.0.1,0.05.0.025.
We can find these by substituting the x values on the table given into the equation; h(x)=1/x.
Answer:
Please see the answer below
Step-by-step explanation:
a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000
b. The number of ways such that the 4 math books remain together
The pattern is as follows: MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM
Where M = Math’s Book and E= English Book.
Number of ways = 4!*8!*4*150= 86400 ways.
c. The number of ways such that math book is at the beginning of the shelf
The number of ways = 6!*4*150 = 432000
d. The number of ways such that math and English books alternate
The number of ways = 150*4!*3! =2160 ways
e. The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.
