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gregori [183]
3 years ago
9

An equilateral triangle is always __________. A. obtuse and isosceles B. acute and isosceles C. scalene D. obtuse

Mathematics
2 answers:
V125BC [204]3 years ago
8 0
 acute b that is what it always is

OLEGan [10]3 years ago
4 0

An equilateral triangle is always B. acute and isosceles

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14 1/6 - 7 1/3 in simplest form <br> Help plz
Firdavs [7]

Answer: 7 -1/6

Step-by-step explanation:

Multiply the denominator and numerator (1/3) by 2 to get 2/6, then do 1/6-2/6, which gives you a negative. Then subtract your whole numbers (14-7) to get 7 -1/6

7 0
3 years ago
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What is the most precise name for quadrilateral ABCD with vertices
Sloan [31]
Square    when you look at the graphed answer
                               
6 0
4 years ago
(2,-4) reflected over the line y=-x
MatroZZZ [7]

Answer:

(2,4)

Step-by-step explanation:

that would be the answer because when you reflect it over the Y axis you are making the y turn into a + or - depending on what it already is if y=-4 when you reflect it over the y it would be y=4

6 0
4 years ago
Can anyone Help me with this problem??
lesya692 [45]

Answer:

9.

Step-by-step explanation:

So basically, we need to find out x. So we know that ZX = YW because ZY and XY are already congruent and it's just adding a three to the value because of YX. We can find out x.

ZX = YW

5x + 17 = 10 - 2x

x = -1

Then we can find out the value of ZX and YW by plugging in x.

5(-1) + 17 = ZX = YW = 12.

Then we see that YX is 3 so...

ZX = ZY + YX

12 = ZY + 3

ZY = 9

So the answer is 9.

4 0
3 years ago
Oordinates of point that divides the join of A (-6, 3) &amp; B (5, -2) in ratio 2 : 3 externally _________.
DedPeter [7]

Answer:

C(x,y) = (-28,13)

Step-by-step explanation:

If co-ordinates of points A(x1,y1) and (x2,y2) and C is the external point that divides the line AB in ratio m:n then,

C (x,y) = \frac{mx_2-nx_1}{m-n} , \frac{my_2-ny_1}{m-n}

Given: A (-6, 3) & B (5, -2) in ratio 2 : 3

Substituting these values in above formula to C(x,y)

C (x,y) = \frac{2\times5-3(-6)}{2-3} , \frac{2(-2)-3(3)}{2-3}

Solving we get

C(x,y) = (-28,13)

5 0
3 years ago
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