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NemiM [27]
2 years ago
13

Need help with this geometric question ASAP, Thank you

Mathematics
1 answer:
Harlamova29_29 [7]2 years ago
6 0

Answer: 15

Step-by-step explanation:

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Heyy I don’t think your pictures are loading it just shows grey
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3 years ago
Two wires
xz_007 [3.2K]

Answer:

am i suppoesd to do number 7 as well?

Step-by-step explanation:

Two wires

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4 0
3 years ago
What is 2,001÷83 pls help i need to check my answer
Nina [5.8K]

Answer:

24.11

Step-by-step explanation:

2001 ÷ 83

83×2= 166

200-166=34 bring down the 1

341 ÷ 83

83×4= 332

341-332=9 add a decimal and bring down a 0

90÷83

90-83=7 bring down another 0

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700÷83

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round to the nearest tenth

24.11

7 0
3 years ago
A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

4 0
3 years ago
In ABC, BC = a = 16, AC = b = 10, and m&lt;C = 22º. Which equation can you use to find the value of c = AB?
AnnyKZ [126]

Answer:

10 times 16 = C

Step-by-step explanation:

c2 = a2 + b2 − 2ab cos C = 162 + 102 − 2(10)(16) cos 22°

5 0
3 years ago
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