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IgorC [24]
3 years ago
13

Ayo i need help on math 3x-45=12

Mathematics
1 answer:
Leto [7]3 years ago
6 0

x=19

add 45 to both sides 3

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Find the measure of the indicated angle.
marusya05 [52]

Answer:

the answer is 96°

Step-by-step explanation:

it is an isosceles triangle as it's 2 sides are equal.

for an isosceles triangle, the angles made by the equal sides with the third side are equal.

and the sum of all angles in a triangle = 180°

42 + 42 + x = 180

x = 180 - 84

X = 96°

4 0
3 years ago
BRAINLIEST ANSWER PLZ HELP THANK YOU
insens350 [35]

For this you should just look at the y value in each ordered pair.

The first one should be no.

second one: yes

third one: no

fourth one: no

4 0
4 years ago
Read 2 more answers
Ok so i know that you want brainleist but pls answer first?!
andrew-mc [135]

Step-by-step explanation:

Expression: \frac{40}{y - 16}

When \: y=20, \: the \: value \: of \: the \: expression \: is  \:  \frac{40}{4}  = 10

5 0
3 years ago
If 300 cubes can fit in the rectangular prism below what is the edge length of each cube
ehidna [41]

The answer is 1/4 inches

7 0
3 years ago
BRAINLIESTTT ASAP! PLEASE HELP ME :)
yan [13]

Answer:

<em>P=1620</em>

<em>Third option</em>

Step-by-step explanation:

<u>Horizontal Asymptotes</u>

A given function is said to have a horizontal asymptote in y=a, if:  

\displaystyle \lim _{x\rightarrow -\infty }f(x)=a

Or,

\displaystyle \lim _{x\rightarrow +\infty }f(x)=a

For the given function, the population of the species of bird is given by :

\displaystyle p(t)=\frac{1620}{1+1.15e^{-0.042t}}

Where t is the time in years. To find the horizontal asymptote, we should compute both limits to check if they exist.  

\displaystyle \lim _{x\rightarrow +\infty }\frac{1620}{1+1.15e^{-0.042t}}=\frac{1620}{1+0}=1620

When t tends to plus infinity, P tends to 1620 .

The second asymptote is computed by:

\displaystyle \lim _{x\rightarrow -\infty }\frac{1620}{1+1.15e^{-0.042t}}=\frac{1620}{1+\infty}=0

When t tends to minus infinity, P tends to zero. Since the domain of P is t\geq 0, this asymptote is not valid, thus our only asymptote is

\boxed{P=1620}

6 0
3 years ago
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