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mestny [16]
3 years ago
10

I need help depratly please

Mathematics
1 answer:
Bess [88]3 years ago
6 0

Answer:

nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnbbbbbbbb

Step-by-step explanation:nbnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

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M. MRO = 127° m. MRP = 87º and m. NRQ = 83<br> Find m NRP
stiks02 [169]

Answer:

don't know answer what is answee

5 0
3 years ago
What are the center and radius of the circle with equation (−)+(−)=?Open x minus 4 close squared . Plus . Open y minus 9 close s
lbvjy [14]

Answer:

Center and radius of the circle of the  equation  {(x - 4)^{2} + (y - 9)^{2} = 1

center = ( 4,9)         Radius = 1

Step-by-step explanation:

Given -

Equation of circle is  (x - 4)^{2} + (y - 9)^{2} = 1

The equation of circle is  represented by

   \boldsymbol{\mathbf{(x - a)^{2} + (y - b)^{2} = r^{2}}}

where a and b are center of circle and radius r

Comparing these equation

then a = 4  and b = 9    and r = 1

3 0
4 years ago
Geometry! ZK, ZL, ZM are perpendicular bisectors of AGHJ. Select two options that are true.please I need help I don’t get it!!!!
nikdorinn [45]

Answer:

Answer B and C

Step-by-step explanation:

Z is the center of the circle passing through H,G,J:

HZ=85=ZG=ZJ ==> ZJ=85 : B is true

HK=KG=136/2=68: C is true.

6 0
3 years ago
T + 3t - 7 = 4t - 7???
kupik [55]
t+3t-7=4t-7\\\\4t-7=4t-7\ \ \ \ \ |subtract\ 4t\ from\ both\ sides\\\\-7=-7-TRUE\\\\infinitely\ many\ solutions;\\each\ number\ is\ a\ root\ of\ the\ equation\\\\t\in\mathbb{R}
6 0
4 years ago
Read 2 more answers
Write any two quadratic equations​.
Thepotemich [5.8K]

To find:

Any two quadratic equations​.

Solution:

The general form of a quadratic equation is:

ax^2+bx+c=0

Where, a,b and c are real number and a is non zero.

We know that a,b,c can take any values but a cannot be 0.

For a=1, b=1,c=1, we get

1x^2+1x+1=0

x^2+x+1=0

One quadratic equation is x^2+x+1=0.

For a=3,b=-1,c=5, we get

3x^2+(-1)x+5=0

3x^2-x+5=0

Therefore, the two quadratic equations are x^2+x+1=0 and 3x^2-x+5=0.

3 0
3 years ago
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