What is the quotient? StartFraction t + 3 Over t + 4 EndFraction divided by (t squared + 7 t + 12)

Question

What is the quotient?

StartFraction t + 3 Over t + 4 EndFraction divided by (t squared + 7 t + 12)

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Eliza 3 months 2022-02-10T10:26:55+00:00 2 Answers 0 views 0

Answers ( )

    0
    2022-02-10T10:28:13+00:00

    The quotient is:

    [tex]\frac{1}{(t+4)^2} \text{ or } \frac{1}{t^2+8t+16}[/tex]

    Solution:

    Given expression is:

    [tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}}[/tex]

    We have to find the quotient

    From given,

    [tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{t+3}{t+4} \times \frac{1}{(t^2+7t+12)} —— eqn 1[/tex]

    [tex]Let\ us\ factor\ (t^2+7t+12)[/tex]

    Split the middle term

    [tex]t^2+7t+12 = t^2 + 3t+4t+12[/tex]

    Group the terms

    [tex](t^2+3t) + (4t+12)[/tex]

    Take the common factor out

    [tex]t(t+3)+4(t+3)[/tex]

    Again take (t+3) as common factor

    [tex](t+3)(t+4)[/tex]

    Substitute in eqn 1

    [tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{t+3}{t+4} \times \frac{1}{(t+3)(t+4)}[/tex]

    Cancel the common factors

    [tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{1}{(t+4)^2}[/tex]

    Therefore, quotient is:

    [tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{1}{(t+4)^2} = \frac{1}{t^2+8t+16}[/tex]

    0
    2022-02-10T10:28:43+00:00

    Answer:

    Option 3!

    Step-by-step explanation:

    Answer on Edge. Took forever to find.

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45:7+7-4:2-5:5*4+35:2 =? ( )