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2 years ago

Point c is located in which quadrant?

Mathematics

2 answers:

Murrr4er [49]2 years ago

7 0

Answer 4

Step-by-step explanation:

Top right 1

Top left 2

Bottom left 3

bottom right 4

vlada-n [284]2 years ago

4 0

Answer:

IV.

Step-by-step explanation:

You need to know that for the x-axis, right means positive values and left means negative values. In this case, C is on the right side of the x-axis, then has a positive x value.

For the y-axis up means positive values and down means negative values. In this case, C is down, so C has a negative y-value.

Now,

I quadrant is x and y positive.

II quadrant is x negative and y positive.

III quadrant is x and y negative.

IV quadrant is x positive and y negative.

Then as C has x positive and y negative, C is in IV quadrant.

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The measures of the angles 7x+17 and 3x-20

Liula [17]

The measures of the two angles are 82.1 and 7.9.

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3 years ago

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Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

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

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

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

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

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

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

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago

The points (–2, 4) and (–2, –2) are vertices of a heptagon. Explain how to find the length of the segment formed by these endpoi

Scrat [10]

Answer:

The segment is 6 units long.

Step-by-step explanation:

The points (–2, 4) and (–2, –2) are vertices of a heptagon. We have to explain how to find the length of the segment formed by these endpoints.

If two points at the ends of a straight line PQ are P( $x_{1},y_{1}$ ) and Q( $x_{2},y_{2}$ ), then the length of the segment PQ will be given by the formula