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2 years ago

Help! need these answers filled in for school asap

Mathematics

1 answer:

vampirchik [111]2 years ago

3 0

Answer:

Given : JKLM is a rectangle.

Prove: JL ≅ MK

Since, by the definition of rectangle all angles of rectangles are right angle.

Thus, In rectangle JKLM,

∠ JML and ∠KLM are right angles.

⇒ ∠ JML ≅ ∠KLM

Since, JM ≅ KL (Opposite sides of rectangles are congruent)

ML ≅ ML ( Reflexive )

Thus, By SAS congruence postulate,

Δ JML ≅ Δ KLM

⇒ JL ≅ MK ( because corresponding parts of congruent triangles are congruent)

Hence proved.

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So;

$f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1$

Evaluating the second function;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}$

simplifying further;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}$

Therefore, we have;

$undefined$

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Step-by-step explanation:

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