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NeTakaya
3 years ago
5

Triangle JKL is rotated 45° counterclockwise using the origin as the center of rotation. On a coordinate plane, triangle J K L h

as points (negative 2, 0), (0, negative 4), (negative 3, negative 5). Which graph shows the location of triangle J’K’L’? On a coordinate plane, triangle J prime K prime L prime has points (negative 1.5, negative 1.5), (3, negative 3), (1.2, negative 5.7). On a coordinate plane, triangle J prime K prime L prime has points (Negative 1.5, 1.8), (negative 3, negative 3), (negative 6, negative 1.2). On a coordinate plane, triangle J prime K prime L prime has points (negative 4, negative 1), (0, negative 2.2), (negative 1.2, negative 5). On a coordinate plane, triangle J prime K prime L prime has points (0, negative 2), (4, 0), (5, negative 3).
Mathematics
2 answers:
Taya2010 [7]3 years ago
6 0

Answer:

the correct answer is a

Step-by-step explanation:

took the edgenuity test and recieved a perfect score

kondaur [170]3 years ago
5 0

Answer:

hey its A

Step-by-step explanation:

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Write an equation in slope intercept form (y=mx+b) for the graph below .
jolli1 [7]

Answer:

Step-by-step explanation:

b would be 0

your slope is rise/run. up one and over three. up one and over three. so 1/3

write it as y=1/3x+0, or y=1/3x

6 0
3 years ago
2y-18x=-26 solve for y
Alekssandra [29.7K]
2y - 18x = -26

Add 18x to both sides.

2y = -26 + 18x

Divide by 2 on both sides.

y = -13 + 9x

y = 9x - 13
8 0
2 years ago
Please show work and thank youuu
Assoli18 [71]

Answer:  6\sqrt{3}

======================================================

Explanation:

Method 1

We can use the pythagorean theorem to find x.

a^2+b^2 = c^2\\\\6^2+x^2 = 12^2\\\\36+x^2 = 144\\\\x^2 = 144-36\\\\x^2 = 108\\\\x = \sqrt{108}\\\\x = \sqrt{36*3}\\\\x = \sqrt{36}*\sqrt{3}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 2

Use the sine ratio to find x. You'll need a reference sheet or the unit circle, or simply memorize that sin(60) = sqrt(3)/2

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(60^{\circ}) = \frac{x}{12}\\\\\frac{\sqrt{3}}{2} = \frac{x}{12}\\\\x = 12*\frac{\sqrt{3}}{2}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 3

Similar to the previous method, but we'll use tangent this time.

Use a reference sheet, unit circle, or memorize that tan(60) = sqrt(3)

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(60^{\circ}) = \frac{x}{6}\\\\\sqrt{3} = \frac{x}{6}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 4

This is a 30-60-90 triangle. In other words, the angles are 30 degrees, 60 degrees, and 90 degrees.

Because of this special type of triangle, we know that the long leg is exactly sqrt(3) times that of the short leg.

\text{long leg} = (\text{short leg})*\sqrt{3}\\\\x = 6\sqrt{3}\\\\

The short leg is always opposite the smallest angle (30 degrees).

3 0
2 years ago
Almonds are a member of the peach family
Alex777 [14]

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6 0
3 years ago
Help plz and check just to make sue what is this
padilas [110]

use cross  products to solve

5* (n-9) = 8*n

distribute

5n-45 = 8n

subtract 5n from both sides

-45 = 3n

divide both sides by 3

-15 = n

3 0
3 years ago
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