Which of the following is an equivalent form of the compound inequality −33 > −3x − 6 ≥ −6?

Mathematics

1 answer:

djyliett [7]2 years ago

6 0

Answer:

$-3x-6 < -33$ and $-3x-6 \geq -6$

Step-by-step explanation:

we have

$-33 > -3x-6 \geq -6$

we know that

Compound inequality can be divided into two inequalities

$-33 > -3x-6$

rewrite

$-3x-6 < -33$

and

$-3x-6 \geq -6$

therefore

An equivalent form of the compound inequality is

$-3x-6 < -33$ and $-3x-6 \geq -6$

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Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for <em>area</em> of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$