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xxMikexx [17]
3 years ago
13

In the following diagram which segments are radii of circle p​

Mathematics
1 answer:
lapo4ka [179]3 years ago
5 0

Answer:

The correct answers would be A,C,D

Step-by-step explanation:

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12x - 2y = -1<br>+ 4x + 6y= -4<br>​
svp [43]

Answer:

x = -7/40 , y = -11/20

Step-by-step explanation:

Solve the following system:

{12 x - 2 y = -1 | (equation 1)

4 x + 6 y = -4 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{12 x - 2 y = -1 | (equation 1)

0 x+(20 y)/3 = (-11)/3 | (equation 2)

Multiply equation 2 by 3:

{12 x - 2 y = -1 | (equation 1)

0 x+20 y = -11 | (equation 2)

Divide equation 2 by 20:

{12 x - 2 y = -1 | (equation 1)

0 x+y = (-11)/20 | (equation 2)

Add 2 × (equation 2) to equation 1:

{12 x+0 y = (-21)/10 | (equation 1)

0 x+y = -11/20 | (equation 2)

Divide equation 1 by 12:

{x+0 y = (-7)/40 | (equation 1)

0 x+y = -11/20 | (equation 2)

Collect results:

Answer:  {x = -7/40 , y = -11/20

6 0
3 years ago
The location of point V is (-3,3). The location of point X is (9,13). Determine the location of point W which is 3/4 of the way
Inga [223]

ANSWER

W(\frac{9}{7},\frac{51}{7} )

EXPLANATION

We want to find the coordinates of the point W(x,y) which divides V(-3,3) and X(9,13) in the ratio m:n=3:4.

The x-coordinate of this point is given by:

x= \frac{mx_2+nx_1}{m + n}

x= \frac{3(9)+4( - 3)}{4 + 3}

x= \frac{21 - 12}{4 + 3}

x= \frac{9}{7}

The y-coordinates is given by;

y= \frac{my_2+ny_1}{m + n}

y= \frac{3(13)+4( 3)}{4 + 3}

y= \frac{39+12}{4 + 3}

y= \frac{51}{7}

Hence

W(\frac{9}{7},\frac{51}{7} )

3 0
3 years ago
SOMEONE PLEASE ACTUALLY HELP MEEE!!!!! Write an example of the Identity Property of Addition using the number 43.
masya89 [10]

Answer:

40 + 3

Step-by-step explanation:

=43. you dont need to use it for 0, even though that's adding too. and multiplying isnt addition.

8 0
3 years ago
How can you use multiplication to help you divide 6,000 by 20?
Anika [276]

Answer:

20 x ___ = 6000

Step-by-step explanation:

Thats how you use multiplication

HOPE THIS HELPS

PLZZ MARK BRAINLIEST

4 0
3 years ago
Read 2 more answers
Solution for: <br> -x + 3y -2z = 19<br> 2x + y - z = 5<br> -3x - y + 2z = -7
STatiana [176]

Answer:

x = -1 , y = 4 , z = -3

Step-by-step explanation:

Solve the following system:

{-x + 3 y - 2 z = 19 | (equation 1)

2 x + y - z = 5 | (equation 2)

-3 x - y + 2 z = -7 | (equation 3)

Swap equation 1 with equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

2 x + y - z = 5 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Add 2/3 × (equation 1) to equation 2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y/3 + z/3 = 1/3 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Multiply equation 2 by 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Subtract 1/3 × (equation 1) from equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

0 x+(10 y)/3 - (8 z)/3 = 64/3 | (equation 3)

Multiply equation 3 by 3/2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

0 x+5 y - 4 z = 32 | (equation 3)

Swap equation 2 with equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+y + z = 1 | (equation 3)

Subtract 1/5 × (equation 2) from equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+0 y+(9 z)/5 = (-27)/5 | (equation 3)

Multiply equation 3 by 5/9:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Add 4 × (equation 3) to equation 2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y+0 z = 20 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Divide equation 2 by 5:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Add equation 2 to equation 1:

{-(3 x) + 0 y+2 z = -3 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(3 x)+0 y+0 z = 3 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Divide equation 1 by -3:

{x+0 y+0 z = -1 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Collect results:

Answer:  {x = -1 , y = 4 , z = -3

6 0
3 years ago
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