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anastassius [24]
4 years ago
13

Write an equation for each:

Mathematics
1 answer:
liraira [26]4 years ago
4 0
1. AB \cong CD
congruent means equal

2. AB \cong BC
bisect means to cut a line segment in two equal parts

Above the AB, CD, AB, and BC there has to be a line. I could not write that because LaTeX doesn't allow it.

3. 2x + 5 = 4x-8
You set both equal to each other

4. PA \cong AT
midpoint means that both line segments are equal

Same thing goes for PA, and AT. You need to draw a line above it.

Hope that helped :)

You might be interested in
9)Solve e/4 – 10 = -20 (CHECK on your graph paper.)
rusak2 [61]

Answer:

9) e=-40 10) k=9

Step-by-step explanation:

9) e/4-10=-20  add 10 on both sides

e/4=-10  Multiply everything by 4

e=-40

10) 0.4k-4+1.6k=14   combine like terms

2k-4=14   Add 4 on both sides

2k=18   divide both sides by 2

k=9


5 0
3 years ago
How many triangles can be constructed with sides measuring 14cm, 8 cm, and 5 cm?
Kryger [21]
 Answer: B

Explanation: No triangle can be formed.

According to the triangle inequality theorem, the sum of any two sides must be greater or equal than the length of the third side.

Here, the sides are 14 cm, 8 cm, and 5 cm.
14cm + 8 cm = 22 cm > 5cm (theorem followed)
14 cm + 5 cm = 19 cm > 8 cm (theorem followed)
8 cm + 5 cm = 13 cm < 14 cm (theorem not followed)

Since, the triangle inequality theorem is not followed, Hence, no triangle can be formed.
8 0
3 years ago
Please answer the questions before 12 am
fomenos
The questions are answered there’s no further more answers
3 0
3 years ago
HELP ME PLEASEEEEEEEEEE
eimsori [14]

Answer:

1. ratio - 1:4 (side to perimeter)

2. ratio - 4:1 (perimeter to side)

3. ratio - 1: 2 (radius to diameter)

Step-by-step explanation:

for 1. I just put 6/24 which simplies to 1/4. I tried the that with the rest and it equalled to 1/4

for 2. put 24/6 which is 4 (in fraction from 4 is 4/1)

for 3. 4/8 which equals to 1/2

4 0
3 years ago
What are the zeroes of y=x^2+14x+40
a_sh-v [17]
y=x^2+14x+40 

y-x^2-14x-40=0 

In general, given a{x}^{2}+bx+c=0, there exists two solutions where

x=  \dfrac{14+2 \sqrt{y+9} }{-2} , \dfrac{14-2 \sqrt{y+9} }{-2} 

x=-7- \sqrt{y+9},-7+ \sqrt{y+9}
3 0
3 years ago
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