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kogti [31]
3 years ago
12

Which of these can not form a perpendicular bisector?

Mathematics
2 answers:
marusya05 [52]3 years ago
6 0
B is the answer

Hope it helps
sdas [7]3 years ago
4 0
B. Is the correct answer. A circle can not form a perpendicular bisector as it is rounded and not straight.
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If 45.78 ÷ 1.7 is written in long division form as long division set up where 17 is on the outside of the division symbol and 45
Nostrana [21]

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1.7 | 45.78

ok....we want to make the 1.7 into a whole number....we do this by moving the decimal 1 space to the right.....and since we move the decimal one space on the divisor, we need to move the decimal 1 space on the dividend.

so ur problem would now be :
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17 | 457.8
7 0
3 years ago
Read 2 more answers
Just keep getting more homework.<br> i really need help
rusak2 [61]

Answer:

1. x = 1, y = -3

2. x = 3, y = 4

Step-by-step explanation:

Here, we want to solve the sets of simultaneous equations;

a) -y = 5x-2

-3y = -5x + 14

substitute 1 into ii

3(5x-2) = -5x + 14

15x - 6 = -5x + 14

15x + 5x = 14 + 6

20x = 20

x = 20/20

x = 1

from;

-y = 5x - 2

y = -5x + 2

y = -5(1) + 2

y = -3

b) -2y = -6x + 10

2y = x + 5

Substitute ii into i

-(x + 5) = -6x + 10

-x - 5 = -6x + 10

-x + 6x = 10 + 5

5x = 15

x = 15/5

x = 3

2y = x + 5

2y = 3 + 5

2y = 8

y = 8/2

y = 4

4 0
3 years ago
Simplify the square root.<br> 6-63
Zina [86]

Answer:

18i\sqrt{x} 7\\

Step-by-step explanation:

the answer is 18i in front if the root square an 7 goes inside it

4 0
3 years ago
Please help me with #16 I am really bad at doing fractions
emmasim [6.3K]
To see if it's a right triangle we need to determine if a²+b²=c²
(\frac{3}{20})² + (\frac{1}{5})² = \frac{1}{4}
\frac{9}{400} + \frac{1}{25} = \frac{1}{16}
\frac{9}{400} + \frac{16}{400} = \frac{25}{400}
\frac{25}{400} = \frac{25}{400}

So it is a right triangle
3 0
3 years ago
Graph the line with a slope of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B3%7D" id="TexFormula1" title="\frac{5}{3}" alt
balu736 [363]
I don’t know sorry please somebody else can help you
8 0
3 years ago
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