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jasenka [17]
4 years ago
13

PLEASE HELP ME!! PLEASE!

Mathematics
1 answer:
juin [17]4 years ago
5 0
1. There is no "h" in f(x) = log(x). The question is nonsense.

2. The log function is defined for arguments greater than zero. You're probably expected to select D, though A and B could also be correct.

3. Same deal. The log function is defined for x > -2, but the domain could be restricted to x > 2. You're expected to select FALSE.

4. Same deal. Choose B, though D is also correct.

5. C is the proper choice. (See discussion of the log function, above.)
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Someone pls help me with this I'm begging you to pls not write whatever and steal points I rlly need help I will mark you as bra
Rashid [163]

Answer:

okay you said '' and steal my points'' so ill do

Step-by-step explanation:

8 0
3 years ago
One third of a number increased by twice the number
ioda
1/3x+3x. Hope that helps! :D
8 0
3 years ago
Explain answer <br> GIVING BRAINIEST
AveGali [126]

Answer:

The slope of the line is 2. To find the slope I marked two points on the line, which were (-3, 0) and (-2, 2). Then I used the \frac{Rise}{Run} technique. Moving up 2 units and to the right 1 unit, which would be represented by \frac{2}{1}.

5 0
2 years ago
Read 2 more answers
Seclect all the sets of numbers that are possible values for x in the inequalit, x &gt; - 8. { -10, -8, -6}. { -8, 0, 8}. { -9,
Sauron [17]

Answer:

{ -6, -4, -2}

Step-by-step explanation:

Since x > - 8, x can be any number greater than - 8.

x = -7, -6, -5, -4,...

BUT, it can't be equal because it's not using this sign \geq

--------------------------------------------------------------------------------

--------------------------------------------------------------------------------

1. { -10, -8, -6} can't be an inequality of x because it has -10, -8.

2.{ -8, 0, 8} can't be an inequality of x because of -8.

3.{ -9, 1, 7} can't be an inequality of x because -9 isn't greater than x.

4 0
3 years ago
What equation is graphed in this figure?
Ksivusya [100]

to get the equation of any straight line, we simply need two points off of it, let's use those two points in the picture below.

(\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}}\implies \cfrac{3}{1}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_2}{5}=\stackrel{m}{3}(x-\stackrel{x_2}{1})

keeping in mind that for the point-slope form, either point will do, in this case we used the second one, but the first one would have worked just the same.

7 0
2 years ago
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