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3 years ago

given the parallel lines cut by. a transversal , determine the value of x . please help due in 30 minutes

Mathematics

2 answers:

Mademuasel [1]3 years ago

7 0

Answer:

x = 20

Step-by-step explanation:

The angles are same side interior angles so the add to 180

3x-10 + 5x+30 = 180

Combine like terms

8x +20 =180

Subtract 20 from each side

8x+20 -20 =180-20

8x = 160

Divide by 8

8x/8 =160/8

x = 20

Tresset [83]3 years ago

3 0

Answer: x = 25

Step-by-step explanation:

When lines are parallel, the same side interior angles are supplementary, meaning they add up to 180 degrees. So to find x, you would add the two expressions and make it equal to 180 degrees.

3x - 10 + 5x + 30 = 180

8x - 20 = 180

8x = 200

x = 25

The answer is x=25

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Hey there! :D

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3 years ago

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If 7y + 42 <14 then which among the following is true

lutik1710 [3]

Answer:

$y$

Step-by-step explanation:

Given

$7y + 42$

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Options are not given; but, the question can still be attempted.

$7y + 42$

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3 years ago

Graph the function y = x3 + x2 – 6x + 5. What are the coordinates, to the nearest tenth, of the relative maximum of the graph? (

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Step-by-step explanation:edge

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3 years ago

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Given the arc length of a circle is 127π centimeters and the subtended angle is 60˚. Find the Circumference of the circle.

Mumz [18]

The circumference of the circle is 762π cm.

<h3>What is length of arc of a circle?</h3>

The length of arc a circle can is determined with the radius and central angle of the arc.

The radius of the circle is calculated as follows;

$L = \pi r \times \frac{ \theta }{180} \\\\
\pi r \theta = 180 L\\\\
\pi r = \frac{180 L}{\theta}$

The circumference of the circle is calculated as follows;

$C = 2\pi r\\\\
C = 2(\pi r)\\\\
C = 2(\frac{180 \ L}{\theta } )\\\\
C = 2(\frac{180 \times 127\pi}{60 } )\\\\
C = 762 \pi \ cm$

Learn more about circumference of a circle here: brainly.com/question/9782777

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2 years ago

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

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3 years ago