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dlinn [17]
4 years ago
12

Let f(x)=x2 - 4x - 7 and g(x)=x-3 fins f(g(x))

Mathematics
1 answer:
iris [78.8K]4 years ago
6 0
Answer: x^2 - 10x + 14

Explanation:

f(g(x))
= f(x-3) Since g(x) = x-3

And f(x) = x^2 - 4x - 7

=> f(x-3) = (x-3)^2 - 4(x-3) - 7
=> f(x-3) = x^2 - 6x + 9 - 4x + 12 - 7
=> f(x-3) = x^2 - 10x + 14




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285.5

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3 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
What is the coordinate?
faltersainse [42]
Remember that the equation of a circle is:
(x - h)^{2} + (y - k)^{2} =  r^{2}
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.

x^{2} - 6x +  y^{2} - 10y = 56

Complete the square. We must do this for x² - 6x and y² - 10y separately.

x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9

Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.

x^{2} - 6x + 9 + y^{2} - 10y = 65

Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.

y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.

(x - 3)^{2}+ y^{2} - 10y + 25= 90

Factor y² - 10y + 25 to get (y - 5)²

(x - 3)^{2}+ (y - 5)^{2} = 90

Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)

Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.
7 0
3 years ago
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