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3 years ago

Consider the figure shown below. Find the value of BL

Mathematics

1 answer:

Gnesinka [82]3 years ago

7 0

Answer:

BL = 2

Step-by-step explanation:

Given:

AL = 3

AC = 10

BC = 9

AL = AM = 3 (tangents drawn form am external point)

CM = AC - AM

CM = 10 - 3

CM = 7

CM = CN = 7 (tangents drawn form am external point)

BN = BC - CN

BN = 9 - 7

BN = 2

BN = BL = 2 (tangents drawn form am external point).

BL = 2

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Step-by-step explanation:

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3 years ago

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masha68 [24]

Answer:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

Step-by-step explanation:

We are given the two functions:

$\displaystyle f(x) = 3x^2 + mx +5 \text{ and } g(x) = nx^2 - 4x -2$

And that:

$\displaystyle h(x) = f(x)\cdot g(x)$

With the given conditions that (1, -40) and (-1, 24) satisfy the new function, we want to determine functions f and g.

First, find h:

$\displaystyle \begin{aligned} h(x) & = f(x)\cdot g(x) \\ \\ & = (3x^2 + mx +5)(nx^2 - 4x -2) \end{aligned}$

Because (1, -40) and (-1, 24) are points on the graph of h, we have that h(-1) = 40 and h(-1) = 24. In other words:

$\displaystyle \begin{aligned} h(1) = -40 & = (3(1)^2 + m(1) +5)(n(1)^2 - 4(1) -2) \\ \\ & = (3 + m +5)(n-4 -2) \\ \\ & = (m+8)(n-6) \\ \\ -40 &= mn-6m+8n-48 \end{aligned}$

And:

$\displaystyle \begin{aligned} h(-1) = 24 & = (3(-1)^2 + m(-1) +5)(n(-1)^2 -4(-1) -2) \\ \\ & = (3 - m +5)(n + 4 -2) \\ \\ & = (-m+8)(n+2) \\ \\ 24 & = -mn -2m + 8n +16 \end{aligned}$

Solve the system of equations. Adding the two equations together yield:

$\displaystyle -16 = -8m+16n - 32$

Solve for either m or n:

$\displaystyle \begin{aligned} -16 & = -8m + 16n - 32 \\ \\ 16 & = -8m + 16n \\ \\ 8m & = 16n - 16 \\ \\ m & = 2n -2\end{aligned}$

Substitute this into one of the two equations above and solve:

$\displaystyle \begin{aligned} -40 & = mn - 6m + 8n - 48 \\ \\ 0 & = (2n-2)n -6 (2n-2) + 8n -8 \\ \\ &= (2n^2 - 2n) + (-12n + 12) +8 n - 8 \\ \\ & = 2n^2 -6n + 4 \\ \\ & = n^2 - 3n + 2 \\ \\ &= (n-2)(n-1) \\ \\ & \end{aligned}$

Therefore:

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Solve for m:

$\displaystyle \begin{aligned}m &= 2n-2 & \text{ or } m & = 2n-2 \\ \\ & = 2(2) - 1 &\text{ or } & =2(1) -2 \\ \\ &= 2 &\text{ or } & = 0 \end{aligned}$

Hence, the values of n and m are either: 2 and 2, respectively; or 1 and 0, respectively.

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2 years ago

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4 years ago

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