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3 years ago

Given the parent functions f(x) = log10 x and g(x) = 3x - 1, what is f(x) g(x)?

Mathematics

1 answer:

Rudiy273 years ago

7 0

Answer:

Option (b) is correct.

The value of $f(x)g(x)=log_{10}x(3x-1)$

Step-by-step explanation:

Given functions $f(x)=log_{10}x$ and $g(x)=3x-1$

We have to find $f(x)g(x)$

We know $f(x)g(x)=f(x) \cdot g(x)$

Thus, $f(x) \cdot g(x)=log_{10}x \cdot (3x-1)$

Thus, solving further we get,

$log_{10}x \cdot (3x-1)=log_{10}x(3x-1)$

Thus, the value of $f(x)g(x)=log_{10}x(3x-1)$

Thus, Option (b) is correct.

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Points X and Z are on a number line, and point Y partitions

abruzzese [7]

Answer:

The coordinate of $Z$ is 13.36.

Step-by-step explanation:

According to the statement, we have the following information:

$\frac{XY}{XZ} = \frac{5}{12}$ (1)

$\frac{YZ}{XZ} = \frac{7}{12}$ (2)

$X = 0.4$ (3)

$Y = 5.8$ (4)

From (2), we have the following expression:

$YZ = \frac{7}{12}\cdot XZ$

$Y-Z =\frac{7}{12}\cdot (X-Z)$

$Y - Z = \frac{7}{12}\cdot X -\frac{7}{12}\cdot Z$

$Y-\frac{7}{12}\cdot X = Z-\frac{7}{12}\cdot Z$

$\frac{5}{12}\cdot Z = Y-\frac{7}{12}\cdot X$

$5\cdot Z = 12\cdot Y-7\cdot X$

$Z = \frac{12}{5}\cdot Y-\frac{7}{5}\cdot X$ (5)

If we know that $Y = 5.8$ and $X = 0.4$ , then the coordinate of $Z$ is:

$Z = \frac{12}{5}\cdot (5.8)-\frac{7}{5}\cdot (0.4)$

$Z = 13.36$

The coordinate of $Z$ is 13.36.

6 0

3 years ago

The solution set for <br> 6a^2-a-5=0

Alex_Xolod [135]

Answer:

a=1 or a=5/6

Step-by-step explanation:

I'm going to attempt to factor 6a^2-a-5

a=6

b=-1

c=-5

Find two numbers that multiply to be a*c and add to be b.

a*c=-30 =-6(5)

b=-1 =-6+5

So replace -a with -6a+5a in the expression we started with

6a^2-6a+5a-5

now we factor by grouping

6a(a-1)+5(a-1)

(a-1)(6a-5)

Now let's solve the equation:

(a-1)(6a-5)=0

So a=1 or a=5/6

4 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$