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2 years ago

What is the graph of the solution to the following compound inequality? –6x – 1 < –25 or 3x + 4 ≤ –5

Mathematics

1 answer:

ivanzaharov [21]2 years ago

4 0

Answer:

$x>4$ or $x \leq -3$

Step-by-step explanation:

Given : $-6x -1 < -25$ or $3x + 4 \leq -5$

Solving first inequality :

$-6x-1$

Add 1 to both sides

$\Rightarrow -6x-1+1$

Divide both sides by 6

$\Rightarrow \frac{6}{6}x>\frac{24}{6}\\\Rightarrow x>4$

Solving second inequality:

$3x + 4 \leq -5$

Subtract 4 from both sides

$\Rightarrow 3x+4-4 \leq -5-4\\\Rightarrow 3x \leq -9$

Divide both sides by 3

$\Rightarrow \frac{3}{3}x \leq \frac{-9}{3}$

$\Rightarrow x \leq -3$

So, $x>4$ or $x \leq -3$

Refer the attached graph

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Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian

777dan777 [17]

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\B=30^o$

Applying the law of sines

Find the measure of angle A

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(A)}=\frac{2}{sin(30^o)}$

$sin(A)=1$

$A=90^o$

Find the measure of angle C

In a right triangle

we know that

$B+C=90^o$ ----> by complementary angles

$B=30^o$

therefore

$C=60^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}$

$c=2\sqrt{3}\ units$

therefore

The dimensions of the triangle are

$A=90^o$

$B=30^o$

$C=60^o$

$a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units$

<em>Second case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\A=30^o$

Applying the law of sines

Find the measure of angle B

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(30^o)}=\frac{2}{sin(B)}$

$sin(B)=0.25$

using a calculator

$B=14.48^o$

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

$A+B+C=180^o$

$A=30^o\\B=14.48^o$

therefore

$30^o+14.48^o+C=180^o$

$C=135.52^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{a}{sin(A)}$

substitute the given values

$\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}$

$c=5.61\ units$

therefore

The dimensions of the triangle are

$A=30^o$

$B=14.48^o$

$C=135.52^o$

$a=4\ units\\b=2\ units\\c=5.61\ units$

see the attached figure to better understand the problem

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