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grigory [225]
3 years ago
15

A polygon on a coordinate grid was transformed using the rule(x,y)>(x-4,y-3).Which of the following describes this transforma

tion
Mathematics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

True

Step-by-step explanation:

Step 1: Find the Coordinates of the Original Triangle

(2, -2)

(6, -3)

(5, 2)

Step 2: Apply the Transformation to the Coordinates

(2-(4),-2+(3))   ->     (-2, 5)

(6-(4),-3+(3))   ->     (2, 0)

(5-(4),2+(3))    ->     (1, 5)

Step 3: Compare the points on the red triangle with the points you got

(-2, 5) = (-2, 5)

(2, 0) = (2, 0)

(1, 5) = (1, 5)

Because the coordinates are the same we can conclude that the triangle was transformed with the rule: (x,y)-> (x-4,y+3)

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Jana mixes 2 cups of black paint and 3 cups of white paint to make a shade of gray paint. How many gallons of black paint will s
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Answer:

I believe you're answers C, (sorry if im wrong)

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3 years ago
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Cho and Jeff each bought a used bike. Jeff's bike was not on sale and cost the same amount as the
antiseptic1488 [7]
The right answer for the question that is being asked and shown above is that: "A. $30." Cho's bike was on sale for $20 less than the original price, and her total cost was half of what Jeff paid. Jeff pay A. $30 <span>for his bike</span>

Jeff = x (original price)
Cho = x-20


8 0
3 years ago
What is the area of the square adjacent to the third side of the triangle?
lana66690 [7]

Answer:

The area of the square adjacent to the third side of the triangle is 11 units²

Step-by-step explanation:

We are given the area of two squares, one being 33 units² the other 44 units². A square is present with all sides being equal, and hence the length of the square present with an area of 33 units² say, should be x² = 33 - if x = the length of one side. Let's make it so that this side belongs to the side of the triangle, to our convenience,

x² = 33,

x = \sqrt{33} .... this is the length of the square, but also a leg of the triangle. Let's calculate the length of the square present with an area of 44 units². This would also be the hypotenuse of the triangle.

x² = 44,

x = \sqrt{44} .... applying pythagorean theorem we should receive the length of a side of the unknown square area. By taking this length to the power of two, we can calculate the square's area, and hence get our solution.

Let x = the length of the side of the unknown square's area -

\sqrt{(44)}^2 = x^2 + \sqrt{33}^2,

x = \sqrt{11} ... And \sqrt{11} squared is 11, making the area of this square 11 units².

3 0
4 years ago
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Can someone please answer this<br><br> - (6m + 7) -6 (-7 - m) = 35
jasenka [17]

Answer:

35=35

Step-by-step explanation:

-(6m+7)-6(-7-m)=35

-6m-7-6(-7-m)=35

5 0
3 years ago
Determine whether the lines L1 and L2 are parallel, skew, or intersecting. L1:x=9+6t,y=12-3t,z=3+9t L2:x=4+16s, y=12-8s, z=16+20
podryga [215]

Answer:

L1 and L2 are skew

Step-by-step explanation:

Since the equation of the line is

L1:x=9+6t,y=12-3t,z=3+9t

L2:x=4+16s, y=12-8s, z=16+20s

then if they intersect each other , they will have both in that point P=(xp , yp ,zp) then

1)9+6t = 4+16s

2) 12-3t =2-8s

3) 3+9t = 16+20s

adding 2*2) to 1)

9+6*t + 24-6t  = 4+16*s + 4-16*s

33 = 8

since this is not possible , the error comes from our assumption that the lines intersect each other

then they are skew or parallel. They are parallel if their corresponding vectors are parallel , that is

L1 (x,y,z) = (9,12,3) + (6,-3,9)*t

L1 (x,y,z) = (4,2,16) + (16,-8,20)*t

then if they are parallel

(16,-8,20)= k*(6,-3,9)

16=6*k

-8 = -3*k

20= 9*k

since there is no k that satisfy for x , y and z simultaneously then L1 and L2 are not parallel

therefore L1 and L2 are skew

3 0
3 years ago
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