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elixir [45]
2 years ago
6

5(F + 4) = 4(f – 6) I REALLY NEED HELP!! please help asap and help me solve it step by stel

Mathematics
1 answer:
Setler [38]2 years ago
5 0
5(F + 4)= 4( f -6)
5f + 20 = 4f -24
-4f. - 4f
F + 20 = -24
- 20. -20
F = -44
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Part F<br> Does this mean that triangle 1 is a right triangle? Why or why not?
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Answer:

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Step-by-step explanation:

5 0
3 years ago
PLZ HELP!!!! WILL MARK BRAINLEST IF CORRECT
OverLord2011 [107]

Answer: the answer is 54

Step-by-step explanation: The error in the expression was that it was multiplied by 1 so it can't be one unless it was also 1 and it's not it is 54.

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Identify the mode of the number 16568
Lynna [10]
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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
I need help pls !!!!!!
swat32

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Trigonometry.

here, we use the Linear pair property of the adjacent angles.

We know that, all the adjacent angles in a linear pair add to get 180°

so we get as,

=> (n+6) +90°+(2n+3) =180°

=> (n+6) +(2n+3) =180-90

=> 3n+9 = 90

=> 3n= 90-9

=> 3n = 81

hence, n = 81/3

=> n = 27°

thus the value of n is 27° .

4 0
3 years ago
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