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garri49 [273]
3 years ago
9

Solutions to this question anyone please?

Mathematics
1 answer:
zhuklara [117]3 years ago
3 0

\frac{a}{x+a} + \frac{b}{x-b} = \frac{a+b}{x+c}\\\frac{a(x-b) + b(x+a)}{(x+a)(x-b)} = \frac{a+b}{x+c} \\\frac{x(a+b)}{(x+a)(x+b)} = \frac{a+b}{x+c}\\x(x+c) - (x+a)(x+b) = 0\\x(c -b -a) = ab\\x = \frac{ab}{c-b-a}

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Define the following terms in your own words in terms of immune response
Tpy6a [65]

Answer:

an antigen is  toxic foreign substance, that induces an immune reaction in the body.

an antibody is a blood protein that is made in retort to the counteracting of a certain antigen.

apoptosis is the death of cells that occur normally and control part of a specific organisims growth.

an autoimmune disease is a condition in which your immune system mistakenly attack your body.

a b-cell is a type of lymphocyte that is responisble for the autoimmunity component in the adaptive immue system.

a t-cell is part of the immue system tht develop from types of stem cells in the bone marrow.

a helper is a type of white blood cells that serves as a specific and important key of immune functions.

a killer is an innate immune cell that shows a strong cytolyic function against physiologically stressed cells such as: tumor cells, and virus infected cells.

a suppressor is a lymphocyte that can suppress the antibody production by other lymphoid cells.

--

i apologize, i can get to more but i am busy.

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Compare the quantity in Column A with the quantity in Column B. Choose the best
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<h3>option :c is correct..ok</h3>

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4 years ago
Pls help me I’ll give out brainliest please dont answer if you don’t know
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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}.

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a traffic survey showed that 1,048 cars passed through an intersection in one hour. which is the best estimate of how many cars
Nikitich [7]
8,384 cars will pass

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