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Umnica [9.8K]
3 years ago
15

Given:Bc biscects ABD mAbd=52° prove: mABC =26°​

Mathematics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

Angle Bisector Theorem

Step-by-step explanation:

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After how many years are the salaries offered by company a and company b the same
12345 [234]
24000 + 3000x = 30000 + 2400x
3000x - 2400x = 30,000 - 24,000
600x = 6000
x = 6000/600
x = 10 years
6 0
3 years ago
You rotate a triangle 90° counterclockwise about the origin. Then you translate its image 1 unit left and 2 units down. The vert
sergiy2304 [10]

Answer:

The coordinates of the original triangle are (2,4). (4,1) and (1,1)

Step-by-step explanation:

About the origin, I have to rotate a triangle 90° counterclockwise. Then I translate its image 1 unit left and 2 units down.

The vertices of the final image are (-5,0), (-2,2), and (-2,-1).

We need to get the vertices of the original triangle.

So, start from a triangle with vertices (-5,0), (-2,2), and (-2,-1) and do the reverse to get the original triangle.

So, we will translate the image triangle by 1 unit right and 2 units up and we will get the triangle with vertices (-5 + 1, 0 + 2), (- 2 + 1, 2 + 2) and (- 2 +1, - 1 + 2)  ≡ (-4,2), (-1, 4) and (-1,1).

Now, we have to rotate this intermediate triangle by 90° clockwise.

Therefore, the coordinates of the original triangle are (2,4). (4,1) and (1,1) (Answer)

{Since, for 90° rotation the coordinates of all the vertices will interchange their location (i.e. x-value to y-value and y-value to x-value) and a negative sign will be added to the y-coordinate of the interchanged coordinates}  

5 0
3 years ago
Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
Alik [6]

Answer:

\alpha=3

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

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

We are given the equation of a circle as

3x^2+3y^2-6\alpha x+12y-3\alpha=0

Note we have corrected it by adding the square to the y. Simplify by 3

x^2+y^2-2\alpha x+4y-\alpha=0

Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

(x-\alpha)^2+(y+2)^2=r^2

We can see that, if r=4, then

\alpha+\alpha^2+4=16

Or, equivalently

\alpha^2+\alpha-12=0

There are two solutions for \alpha:

\alpha=-4,\ \alpha=3

Keeping the positive solution, as required:

\boxed{\alpha=3}

8 0
3 years ago
Simplify the rational expression. State any restrictions on the variable.
ExtremeBDS [4]

we are given

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}

Firstly, we will factor numerators and denominators

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

n^2-6\neq 0

n\neq -+ \sqrt{6}

we can simplify it

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}

we know that denominator can not be zero

n^2-3\neq 0

n\neq -+ \sqrt{3}

so, option-B.......Answer

3 0
3 years ago
How do you write 2sin5cos5 as a single trigonomic ratio?
erastova [34]

Answer:

ertyui

Step-by-step explanation:

6 0
3 years ago
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