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3 years ago

The domain of F(x)=log[b]x is the set of all real numbers. True or false?

Mathematics

1 answer:

tia_tia [17]3 years ago

6 0

$f(x)=log_bx\\\\D=\mathbb{R}\ is\ false.\\\\correct\ domain:D:x\in\mathbb{R^+}\ and\ b\in\mathbb{R^+}\ and\ b\neq1.$

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How do you work this out?

Ira Lisetskai [31]

I'm guessing you find the length of the hypotenuse using the pythagorean theorem.

a² + b² = c² a and b are the legs and c is the hypotenuse

15² + 8² = c²

225 + 64 = c²

289 = c²

√289 = √c²

17 = c

hope that helps, God bless!

3 0

3 years ago

Read 2 more answers

Plz Help!!!! <br>Solve for x<br>3y = 4x+ 1

astraxan [27]

Answer:

Step-by-step explanation:

STEPS FOR SOLVING LINEAR EQUATION

3y=4x+1

Swap sides so that all variable terms are on the left hand side.

4x+1=3y

Subtract 1 from both sides.

4x=3y−1

Divide both sides by 4.

3y−1

Dividing by 4 undoes the multiplication by 4.

3y−1

7 0

3 years ago

Solve for y. show work. 4(y+3)−2y+7

olga_2 [115]

Dude I don’t feel like working

5 0

2 years ago

Read 2 more answers

2. What is the effective annual rate of an investment that pays 6% for 5 years, compounded semiannually?

Ede4ka [16]

Answer:

effective annual rate is 6.16 %

Step-by-step explanation:

given data

rate = 6 % = 0.06

time 5 year = 10 semi annually

to find out

effective annual rate

solution

we know formula for annual effective rate of interest is

rate of interest = $(1+ r/n)^{n}$ -1

put here all value

rate of interest = $(1+ 0.06/10)^{10}$ -1

rate of interest = 0.061646

so effective annual rate is 6.16 %

8 0

3 years ago

Look at the graph shown below:

ElenaW [278]

So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-4}{-4-0}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

once you get the slope and solve for "y", that'd be the equation of the line.

7 0

2 years ago