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2 years ago

Find the coordinates of the orthocenter of triangle ABC with vertices A (2, 0), B (2, 4), and C (6, 0).

Mathematics

1 answer:

ss7ja [257]2 years ago

8 0

9514 1404 393

Answer:

(2, 0)

Step-by-step explanation:

The orthocenter of a right triangle is the vertex where the right angle is located. Here, that is vertex A.

The orthocenter is (2, 0).

The orthocenter is the point where the altitudes intersect.

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HELP PLEASE!! Identify the solution set of the inequality using the given replacement set. x < –2; {–5, –2.6, –2, –0.8, 1, 1.

UNO [17]

X< -2 means that x must be smaller than -2, so something like -3, -4, -5, etc

The only solution set that fits this is A.

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3 years ago

Solve for y<br> 6 = 4x + 9y

crimeas [40]

Answer:

y=.44-.66

Step-by-step explanation:

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2 years ago

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Anderson has $50 in his savings account. He deposits $5 every week. His father also deposits $20 into the account every time And

Sav [38]

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3 years ago

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Quadrilateral ABCD has the following vertices.

Marizza181 [45]

A(-5,5)
B(4,5)
C(2,0)
D(-5,-2)

AB,BC,CD,DA

AB = [4-(-5)),5-5]=[9,0]
Lenght $AB= \sqrt{9^2+0^2}= \sqrt{81} =9$

BC = [2-4,0-5]=[-2,-5]
Lenght $BC=\sqrt{(-2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}$

CD = [-5-2,-2-0]=[-7,-2]
Lenght $CD=\sqrt{(-7)^2+(-2)^2}=\sqrt{49+4}=\sqrt{53}$

DA =[-5-(-5),-2-5]=[0,-7]
Lenght $DA=\sqrt{0^2+(-7)^2}=\sqrt{49}=7$

sorted from longest to shortest:
AB, CD,DA,BC
$\sqrt{81}, \sqrt{53}, \sqrt{49}, \sqrt{29}$

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sp2606 [1]

Answer:

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