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2 years ago

PLZ HELP,ANSWER WHAT IS IN THE PICTURE

Mathematics

1 answer:

MrMuchimi2 years ago

3 0

Answer: X'(-3, -2), Y'(-5, 1), and Z'(2, -3)

Step-by-step explanation:

Upon reflection across the x-axis, the x-coordinates remain the same while the signs of the y-coordinates flip. So, the coordinates will be X(3, -2), Y(5, 1), and Z(-2, -3).

Upon reflection across the y-axis, the signs of the x-coordinates will flip while the signs of the y-coordinates remain the same. So, the coordinates will be X′(-3, -2), Y′(-5, 1), and Z′(2, -3).

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Find the value of x and y <1=(10x) <2=(10y+20) <3 =100

Rufina [12.5K]

Answer:

x = 10

y = 6

Step-by-step explanation:

We can use the vertical angle theorem to solve for most of this problem. We know that 1/3 will have the same measure and so will 4/2. We can create an equation to solve for 1 and 3.

10x = 100

x = 10

We aren't given angle 4 so we are going to have to solve that ourselves. We can create an equation knowing that 4 and 3 will create a supplementary angle.

100 + z = 180

z = 80

Now, that we have angle 4, we know that angle 2 will have the same measure because they are vertical angles. Now to find the value of y, we can take the angle measure and solve for it.

10y + 20 = 80

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3 years ago

Read 2 more answers

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

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