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2 years ago

20 points!! Solve this system of lineal equations.separate the x- and y- values with a comma

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

7 0

Solve the following system:{12 x = 54 - 6 y | (equation 1)-17 x = -6 y - 62 | (equation 2)
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1

Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.

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2 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

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2 years ago

Evaluate 0.53n + 5 when n= 3 and when n= -3

Dahasolnce [82]

Answer:

1.59&-1.59

Step-by-step explanation:

.53(3)+5=1.59

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2 years ago