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3 years ago

Use these functions to evaluate the following problems: f(x) = x, g(x) = x - 3, h(x) = x^2 - 9, k(x) = 2x

Mathematics

1 answer:

S_A_V [24]3 years ago

8 0

Answer:

(hk)(x) = 2x³-18x

(k/f)(x) = 2

(h/g)(x) = x+3

Step-by-step explanation:

(hk)(x)

(x² - 9)(2x)

2x³ - 18x

(k/f)(x)

2x/x

(h/g)(x)

(x² - 9)/(x - 3)

[x² - 3²]/(x - 3)

(x + 3)(x - 3)/(x - 3)

x + 3

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What is the value of x in the proportion below? 57=x21

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the first term and the common ratio third term =6 and seventh term= 96

mamaluj [8]

Answer:

First term a₁ = 3/2 and common ratio r = 2

Step-by-step explanation:

We need to find the first term and common ratio while we are given third term = 6 and seventh term = 96

Since common ratio is required so, the sequence is geometric sequence

The formula used is: $a_n= a_1r^{n-1}$

We are given: third term = 6 i,e

$a_3=a_1r(3-1)\\6=a_1r^2----eq(1)$

Seventh term = 96

$a_7=a_1r(7-1)\\96=a_1r^6----eq(2)$

Dividing eq(2) and eq(3)

$\frac{96}{6}=\frac{a_1r^6}{a1r2}\\16=\frac{r^6}{r^2}\\16=r^{6-2}\\=> r^4=16\\Taking \ fourth \ root\\ r=2,-2,2i,-2i\\ \ We \ will \ consider \ only \ positive \ value \ of \ r \ i.e \ \textbf{ r= 2}$

So, Common Ratio r = 2

Finding First term using eq(1)

$a_3=a_1r^2\\6=a_1(2)^2\\6=a_1(4)\\a_1= \frac{6}{4}\\a_1= \frac{3}{2}$

So, First term a₁ = 3/2 and common ratio r = 2

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3 years ago

PLS HELP how many solutions does this system of equations below have? y=8x+4 y=8x+3

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3 years ago

What is 75%,1/6,3/8,and 0.625 from least to greatest

Naya [18.7K]

Convert each term in decimal form. This is avoid confusion and for uniformity and ease in comparing the value of each term.

75%   is   0.750
1/6   is   0.167
3/8   is   0.375
0.625 is   0.625

From least to greatest, the order is:

0.167 , 0.375 , 0.625 , 0.750

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3 years ago

6, 1, -4, -9,...

kykrilka [37]

Texaschic nailed it when saying that it's arithmetic

The recursive way to write this is to say

$a_1 = 6$
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The $a_n$ portion is the nth term while $a_{n-1}$ is the term just before the nth term

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3 years ago