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3 years ago

If you are given the graph of h(x) = log. x, how could you graph m(x) = log2(x+3)?

Mathematics

1 answer:

Agata [3.3K]3 years ago

3 0

Answer:

Option (d).

Step-by-step explanation:

Note: The base of log is missing in h(x).

Consider the given functions are

$h(x)=\log_2x$

$m(x)=\log_2(x+3)$

The function m(x) can be written as

$m(x)=h(x+3)$ ...(1)

The translation is defined as

$m(x)=h(x+a)+b$ .... (2)

Where, a is horizontal shift and b is vertical shift.

If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.

If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.

On comparing (1) and (2), we get

$a=3,b=0$

Therefore, we have to translate each point of the graph of h(x) 3 units left to get the graph of m(x).

Hence, option (d) is correct.

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I don't understand how to do this problem.

Delicious77 [7]

Answers:

x = 4
EF = 14
CF = 7
EC = 7

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

Work Shown:

C is the midpoint of segment EF. This means that EC = CF. In other words, the two pieces are congruent.

Use substitution and solve for x
EC = CF
5x-13 = 3x-5
5x-13+13 = 3x-5+13
5x = 3x+8
5x-3x = 3x+8-3x
2x = 8
2x/2 = 8/2
x = 4

Now that we know that x = 4, we can use this to find EC and CF

Let's compute EC
EC = 5x - 13
EC = 5*x - 13
EC = 5*4 - 13 ... replace x with 4
EC = 20 - 13
EC = 7

Let's compute CF
CF = 3x - 5
CF = 3*x - 5
CF = 3*4 - 5 ... replace x with 4
CF = 12 - 5
CF = 7

As expected, EC = CF (both are 7 units long).

By the segment addition postulate, we can say EC+CF = EF

EC+CF = EF
EF = EC+CF
EF = 7+7
EF = 14

8 0

3 years ago

I am very stuck with this

Bingel [31]

Answer:

see explanation

Step-by-step explanation:

To multiply the vector by a scalar, multiply each of the elements by the scalar.

To add 3 vectors add the corresponding elements of each vector

2a + 3b + 4c

= 2 $\left[\begin{array}{ccc}2\\3\\\end{array}\right]$ + 3 $\left[\begin{array}{ccc}-4\\1\\\end{array}\right]$ + 4 $\left[\begin{array}{ccc}3\\2\\\end{array}\right]$

= $\left[\begin{array}{ccc}4\\6\\\end{array}\right]$ + $\left[\begin{array}{ccc}-12\\3\\\end{array}\right]$ + $\left[\begin{array}{ccc}12\\8\\\end{array}\right]$

= $\left[\begin{array}{ccc}4-12+12\\6+3+8\\\end{array}\right]$

= $\left[\begin{array}{ccc}4\\17\\\end{array}\right]$

8 0

3 years ago

Is 1/3 and 2/6 equivalent

Helga [31]

Yes because 2/6 divided in half is 1/3

8 0

3 years ago

Read 2 more answers

A designer of carved lawn figures needs to reduce the size of a statue by making each linear dimension 40% of the original. If a

Leya [2.2K]

Answer:

N x (0.4)^3

Step-by-step explanation:

8 0

3 years ago

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s

WINSTONCH [101]

Answer: $\bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}$

<u>Step-by-step explanation:</u>

There are 3 conditions that must be satisfied:

f(x) is continuous on the given interval
f(x) is differentiable
f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2 [0, 3]

1. There are no restrictions on x so the function is continuous $\checkmark$

2. f'(x) = 3x² - 2x - 6 so the function is differentiable $\checkmark$

3. f(0) = 0³ - 0² - 6(0) + 2 = 2

f(3) = 3³ - 3² - 6(3) + 2 = 2

f(0) = f(3) $\checkmark$

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1$

Only one of these values (1.8) is between 0 and 3.

5 0

2 years ago