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Rufina [12.5K]
3 years ago
10

brainliest for correct answer please help Find the perimeter of a triangle with the side lengths of 3X, (X-1), and 2(X+4).

Mathematics
1 answer:
Ulleksa [173]3 years ago
5 0

Answer:

6x+7

Step-by-step explanation:

You might be interested in
If x is an acute angle and tan x = 5,
lyudmila [28]
We have that
tan x=5

we know that
tan x =sin x/ cos x
sin x/ cos x=5------> squaring-----> sin² x/cos² x=25
remember that
sin² x+cos² x=1-----> cos² x=1-sin² x
substitute
sin² x/[1-sin² x]=25
sin² x=25-25*sin² x
26*sin² x=25
sin² x=25/26
sin x=5/√26

cos² x=1-(5/√26)²----> 1-(25/26)----> 1/26
cos x=1/√26

the answers are
sin x=5/√26
cos x=1/√26

4 0
3 years ago
Find the value of x in the triangle shown below.<br> 8<br> 2<br> Answer?
Lyrx [107]
The answer is -2 because I already took this
6 0
3 years ago
The ratio of koalas to gum trees is 1 : 3. There are 252 gums trees now. Plans are being made to increase this number to 345. Fi
Advocard [28]
I would be happy to help!

To start off you need to find the amount of koalas there are at the current number of trees by using the ratio one to 3 which means there is one koala for every 3 gum trees. So you need to devide 252 by 3 giving you the answer of 84 koalas currently. Now you nees to find how many koalas there are for the future 345 trees based on the same ratio 1 to 3. This would give you 115 koalas to be expected in the future. Finally to find the amount of koalas increaed you need to subtract 84 from 115 giving you a final answer of 31 koalas.
7 0
4 years ago
Plz answer ASAP I don't understand it easy math
Rashid [163]

Answer:

its not there

Step-by-step explanation:

because i saw it

3 0
3 years ago
For the given function, find the vertical and horizontal asymptote(s) (if there are any). f(x) = the quantity x squared plus fou
Vika [28.1K]

Answer:

Vertical asymptote at x= 6

Oblique asymptote at y=x+10

Step-by-step explanation:

f(x)= \frac{x^2+4x-3}{x-6}

To find out vertical asymptote , we take the denominator and set it =0

x-6=0 , so x=6

Vertical asymptote at x= 6

The degree of numerator is 2  and the degree of denominator is 1.

Here,  the degree of numerator is higher than the degree of denominator. So there will be a slant  or oblique asymptote. we can find it by long division.

                                  x + 10      

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

      x-6              x^2  +  4x  -3

                         x^2  - 6x

                     -------------------------------(subtract)

                                  10x    - 3

                                  10x   - 60

                        -----------------------------(subtract)

                                           57

Oblique asymptote at y=x+10



3 0
4 years ago
Read 2 more answers
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