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-Dominant- [34]
4 years ago
14

Help me please, if u are able to thank you if not then what should I do?

Mathematics
1 answer:
sesenic [268]4 years ago
5 0
Put a ruler Measuring side AB then post again.
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3kg of rump steak costs 42 pounds adel buys 4kg of rump steak work out how much adel pays
12345 [234]

Answer:

I think it's 56 pounds but I could be wrong

3 0
3 years ago
7+c> 33; c> 25 ? whats the answer
nataly862011 [7]
The answer will be any number greater than or equal to 27. Hope this helps :)
4 0
4 years ago
Read 2 more answers
Which statement shows the associative property of addition?
DaniilM [7]
<span>associative property of addition

answer is 
</span><span>(u + 7) + 13 = u + (7 + 13)</span>
8 0
3 years ago
What are the horizontal and vertical asymptotes of r(x)=(6x+1)/(16x^2)+1?
Mazyrski [523]
Horiz. asy.:  experiment with letting x grow larger and larger (without bound).  Eventually the given expression will approach some limiting value.

If we continue to increase x in <span>(6x+1)/(16x^2), the fraction will approach 0.  Try it!  

If we continue to increase x in </span><span>(6x+1)/(16x^2) + 1, the expression will approach 0 + 1, or 1.
</span>
Which one did you mean?

Thus, the horiz. asy. would be y=0 or y=1. 

Vertical asy.:  Identify the x-value, if any, at which the denominator of this fraction = 0.

If you meant <span>(6x+1)/(16x^2)                 (without the +1), the fraction is undefined at x=0.  Thus, the vert. asy. is the vertical line x=0.</span>
4 0
3 years ago
Please hurry it’s timed
pickupchik [31]

Answer:

x = -π

Step-by-step explanation:

An asymptote line is a line that the graph does not intercept or pass through but rather approaches it. An asymptote line also is considered to be a line of undefined value of function because if an asymptote is x = a then f(a) is undefined.

The question gives a function y = csc(x) which is in another form, y = 1/sin(x).

We have to find the value of x that turns the function in undefined and that x-value will be your asymptote.

From a unit circle, we know that sin(-π) = -sin(π) = 0. Since sin(-π) = 0 —> y = csc(-π) = 1/sin(-π) = <u>1/0</u>

Now we find out that at x = -π, csc(x) is not defined and hence, an asymptote is x = -π

8 0
2 years ago
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