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2 years ago

GEOMETRY QUESTION, HELP PLEASEEE.

Mathematics

1 answer:

choli [55]2 years ago

6 0

Answer:

The pre-image C is (2, 5)

Step-by-step explanation:

If the point (x, y) rotated about the origin by angle 90° clockwise, then its image is (y, -x)

If the point (x, y) rotated about the origin by angle 180° clockwise, then its image is (-x, -y)

If the point (x, y) rotated about the origin by angle 270° clockwise, then its image is (-y, x)

∵ Point C was rotated 270° clockwise around the origin

∵ C' = (-5, 2)

→ By using the 3rd rule above

∵ The image of the point (x, y) is (-y, x)

∴ (-y, x) = (-5, 2)

→ That means -y = -5 and x = 2

∴ x = 2

∴ -y = -5

→ Divide both sides by -1

∴ y = 5

∴ The pre-image C = (2, 5)

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Four interior angles of a pentagon measure 156°, 72°, 98°, and 87°. What is the measure of the final interior angle?

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127 degrees

Step-by-step explanation:

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In △ABC, m∠A=44°, m∠B=48°, and a=25. Find c to the nearest tenth.

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Step-by-step explanation:

We use the sine rule to find the missing sides as follows;

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3 years ago

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

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$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

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We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

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We can select any random value from this interval and plug that in inequality

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we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

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3 years ago

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Step-by-step explanation:

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