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Alenkinab [10]
3 years ago
13

What is the image of point (-5,-7) for mapping (x,y) , (x-4, y-4)

Mathematics
1 answer:
Molodets [167]3 years ago
8 0
Answer:
(-9, -11)

Explanation:
We are given that the original point (x,y) is (-5,-7) and that the image point is
(x-4, y-4)

Now, let's check the x-coordinate:
The original coordinate was x and the image one is x-4.
This means that we can obtain the value of the image x-coordinate by subtracting 4 from the original x-value. 
We know that the original x-value is -5, this means that:
x-coordinate of image = -5-4 = -9

Now, let's check the y-coordinate:
The original coordinate was y and the image one is y-4.
This means that we can obtain the value of the image y-coordinate by subtracting 4 from the original y-value. 
We know that the original y-value is -7, this means that:
y-coordinate of image = -7-4 = -11

Based on the above, the image would be at (-5, -11)

Hope this helps :)
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What is the arc measure of BDC in degrees?<br> (4k + 159)<br> P<br> (2k + 153)
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<h2>Explanation:</h2><h2></h2>

The diagram is missing but I'll assume that the arc BDC is:

(4k + 159)^{\circ}

And another arc, let's call it FGH. measures:

(2k + 153)^{\circ}

If those arc are equal, then this equation is true:

(4k + 159)^{\circ}=(2k + 153)^{\circ} \\ \\ (4k + 159)=(2k + 153) \\ \\ \\ Solving \ for \ k: \\ \\ 4k-2k=153-159 \\ \\ 2k=-6 \\ \\ k=-\frac{6}{2} \\ \\ k=-3

Substituting k into the first equation:

\angle BDC=(4(-3)+159)^{\circ} \\ \\ \angle BDC=(-12+159)^{\circ} \\ \\ \boxed{\angle BDC=147^{\circ}}

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3 years ago
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{362 – [63 + (48 ÷ 2) x 2]} + 3(9 +4)
Julli [10]

Answer:

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Step-by-step explanation:

<u>Solving with one operation at each step:</u>

  • {362 – [63 + (48 ÷ 2) x 2]} + 3(9 +4) =
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What is the answer to 18x-(8x-7)=67
Elza [17]

18x-8x+7=67

10x+7=67

10x=60

x=6

Check Your Work

18(6)-48+7=67

108-48+7=67

60+7=67

Hope this helps...

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Maksim231197 [3]

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The correct answer would be 9

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