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3 years ago

If f(x) = x/2 -2 and g(x) = 2x+ +x= 3, find (f + g)(x).

Mathematics

1 answer:

notka56 [123]3 years ago

6 0

Answer:

.❥︎ᴀᴀɴᴋʜ ᴜᴛʜɪ ᴍᴏʜᴀʙʙᴀᴛ ɴᴇ ᴀɴɢᴅᴀɪ ʟɪ ,

.ᴅɪʟ ᴋᴀ sᴀᴜᴅᴀ ʜᴜᴀ ᴄʜᴀɴᴅɴɪ ʀᴀᴀᴛ ᴍᴇ ❣︎

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Determine whether the dilation from Figure A to Figure B is a reduction or an enlargement

katrin [286]

Answer:

see explanation

Step-by-step explanation:

Since figure B is smaller than Figure A then Figure B is a reduction.

To find the scale factor, calculate the ratio of corresponding sides, image to original.

Using the base lines of both triangles, then

scale factor k = $\frac{2}{6}$ = $\frac{1}{3}$

5 0

3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

Dina took 12 hours to read a 600 page book What was her page per minute rate?

jok3333 [9.3K]

50 divide to find the answer

4 0

2 years ago

Penny reads 11 pages in 1/3 hour. How many pages does she read per hour?

dolphi86 [110]

Answer:

33 pages per hour

Step-by-step explanation:

11 pages=1/3 hour

22 pages=2/3 hour

33 pages= 3/3 hour or 1 hour

5 0

3 years ago

A $32 item is marked as 20% off. What is the sale price?

Natasha_Volkova [10]

$25.60

20%=.2
since it's 20% OFF, you want to get rid of .2 of the 32.
so 1-.2=.8
and .8 x 32 = 25.60

5 0

3 years ago

Read 2 more answers