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ozzi
2 years ago
10

Suppose it is known that for a given differentiable function y=f(x), its tangent line (local linearization) at the point where a

=−4 is given by T(x)=5−7(x+4). What must be the values of f(−4) and f′(−4)?
Mathematics
1 answer:
AleksandrR [38]2 years ago
6 0

Answer:

y(-4) = 5

y'(-4) = -7

Step-by-step explanation:

Hi!

Since the tangent line T and the curve y must coincide at x=-4

y(-4) = T(-4) = 5

On the other hand, the derivative of the curve evaluated at -4 y'(x=-4) must be the slope of the tangent line. Which inspecting the tangent line T(x) is -7

That is:

y'(-4) = -7

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Elina [12.6K]

Answer:

y=4

Step-by-step explanation:

2(y+1)-2=4+y

2y+2-2=4+y

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3 years ago
5x+8=3x-6 plz help me asap
Ray Of Light [21]

Answer:

x = -7

Step-by-step explanation:

5x+8=3x-6

Subtract 3x from each side

5x-3x+8=3x-3x-6

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vovikov84 [41]

Factorize the quadratic.

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We have |ab| = |a||b|, so

|2x^2 + 5x + 3| = |2x + 3| |x + 1| > 0

Now, both |2x+3|\ge0 and |x+1|\ge0 (since the absolute value of any number cannot be negative), so we just need to worry about when the left side is exactly zero. This happens for

(2x + 3) (x + 1) = 0 \implies 2x+3 = 0\text{ or }x+1 = 0 \\\\ \implies x = -\dfrac32 \text{ or } x = -1

So the solution to the inequality is the set

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2 years ago
Given the right triangle MLK and PLN below ; PL=15, KN=4 and cos(L)= (4/5). Find the length of KM
icang [17]
We can answer the question above by using the trigonometric functions and the concept of similar triangles. 
Given that cos (L) is 4/5. The length of sides PN and NL can be solved. Please see solution below.
                            4/5 = NL / 15   ;    NL = 12
Solving for PN,
                                 PN² + 12² = 15²
The value of PN is 9.
Then, using the concept of similar triangles,
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Substituting the known values,
                                 4/(4 + 12) = 9 / KM
The value of KM is 36. 
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3 years ago
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