Can someone please help I really don’t understand this please explain.

Mathematics

2 answers:

marta [7]3 years ago

4 0

Answer:

The answer is option D.

Step-by-step explanation:

You have to multiply it :

h(x) × g(x) = (h×g)(x)

(h×g)(x) = (x² + 2x)(-3x + 4)

(h×g)(x) = -3x³ + 4x² - 6x² + 8x

(h×g)(x) = -3x³ - 2x² + 8x

Juliette [100K]3 years ago

4 0

Answer:

Step-by-step explanation:

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$\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6$

therefore
$D:x\in\left< -\dfrac{5}{4};\ 6\right>$

$\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs$
$9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D$
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Need help please :) thanks

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K’ (-3,-3)
L’(1,-1)
M’(-1,-5)
N’(-5,-7)

D’(0,4)
E’(4,3)
F’(2,-5)
G’(-2,-4)

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