3 years ago

Solving linear systems with graphing

1 answer:

6 0

Answer:

All you have to do is graph both the linear functions on a calculator or on paper and the point where both lines intersect is the solution. So that means if it intersects at a point, your solution would be a coordinate.

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(2) Using the distance formula, d = √(x2 - x1)2 + (y2 - y1)2, what is the distance between point (-2, 2) and point (4, 4) rounde

monitta

Using the distance formula, $d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2 } \\$ , what is the distance between point (-2, 2) and point (4, 4) rounded to the nearest tenth?

Using the points given, plug them into the equation.

$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2 } \\d = \sqrt{((4) - (-2))^2 + ((4) - (2)^2 } \\d=\sqrt{(4+2)^2 + (4-2)^2}\\d=\sqrt{(6)^2 + (2)^2} \\d=\sqrt{36+4} \\d=\sqrt{40} \\$

Plug this into a calculator and you get 6.32455532

Since you only need it up to the tenth (0.1), round up 6.32

Since two is lower than four, we drop it.

Therefore, the distance between points (-2, 2) and (4, 4) is 6.3

Hope this helps ^w^

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3 years ago

Please answer! will give brainliest!!!

GalinKa [24]

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The answer is the last one

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Find the surface area of the cylinder.Round your answer to the nearest tenth.

Delvig [45]

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Step-by-step explanation:

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(-8r^2s^-8)(-3r^-7s^8)

sertanlavr [38]

24r^−5s^0 would be your answer

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In what quadrant will θ terminate, if sin θ is positive and sec θ is negative?

kow [346]

Sin x is possitive in two quadrants :

quadrant I and quadrant II

in quadrant I, cos x is positive

in quadrant II , cos x is negative

so i think the answer is : quadrant II

hope this helps

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3 years ago