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MaRussiya [10]
3 years ago
8

What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?

Mathematics
1 answer:
Sati [7]3 years ago
3 0

<u>Answer-</u>

<em>The coordinates of the orthocenter of △JKL is (-4, 8)</em>

<u>Solution-</u>

The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

For a right angle triangle, the vertex at the right angle is the orthocentre of the triangle.

Here we are given the three vertices of the triangle are  J(-4,-1), K(-4,8) and L(2,8)

If the triangle JKL satisfies Pythagoras Theorem, then triangle JKL will be a right angle triangle.

Applying distance formula we get,

JK^2= (-4+4)^2+ (8+1)^2=0+81=81\\\\KL^2= (-4-2)^2+ (8-8)^2=36+0=36\\\\JL^2= (-4-2)^2+(8+1)^2=36+81=117

As,

\Rightarrow 117=81+36

\Rightarrow JL^2=JK^2+KL^2

\Rightarrow \text{JKL is a right angle triangle}

\Rightarrow \angle K=90^{\circ}

Therefore, the vertex at K (-4, 8) is the orthocentre.

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Answer:

A. The greatest one year increase in tuition ocurred from 1997 to 1998.

Step-by-step explanation:

If you look at the graph, at the points above 1997 and 1998, you'll see that the dots are separated by two lines, not just one.

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3 years ago
There are currently 18 pit bulls at the pound. Of the 18 pit bulls, four have attacked another dog in the last year. Joe, a memb
laila [671]

Answer: 0.7748

Step-by-step explanation:

Given : Number of bit bulls at the pound = 18

Number of pit bulls have attacked another dog in the last year =4

The proportion of pit bulls have attacked another dog in the last year:p=\dfrac{4}{18}\approx0.22

Number of the pit bulls selected = 6

The probability of that none of the pit bulls in Joe's group attacked another dog last year :  P(0)=(1-0.22)^6=0.225199600704\approx0.2252

By using binomial , the probability that at least one of the pit bulls in Joe's group attacked another dog last year is given by :-

P(x\geq 1)=1-P(x

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3 years ago
The sum of two times a number and -2 is at least 8<br><br> how can I solve it plz help
Bogdan [553]
To solve, your equation would be 2x - 2 >/= 8. You add 2 to both sides, getting 2x >/= 10. Then divide by two on both sides, your answer would be x >/= 5. (x is greater than or equal to 5).
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3 years ago
Without multiplying all the terms, verify these equalities. a) 10!= 6!7! b) 10!=7!5!3! c) 16!= 14!5!2! d) 9!=7!3!3!2!
Romashka-Z-Leto [24]

Answer:

a)10! = 6×5×4×3×2 ×7!

10! = 6! 7! (Verified)

b)10! = 3! × 5! × 7! = 7!5!3! (Verified)

c)16! = 2! × 5! × 14!

16! = 14!5!2! (Verified)

d)9! = 3! × 3! ×2! × 7!

9! = 7!3!3!2! (Verified)

Step-by-step explanation:

a) 10!= 6!7!

10! = 10×9×8×7!

Reducing to their lowest factors

10! = (2×5)×(3×3)×(2×2×2) × 7!

Rearranging the factors, we have;

10! = (2×3)×(5)×(2×2)×(3)×2 × 7!

10! = 6×5×4×3×2 ×7!

10! = 6! 7! (Verified)

b) 10!=7!5!3!

10! = 10×9×8×7!

10! = (2×5)×(3×3)×(2×2×2) × 7!

Rearranging the factors, we have;

10! = (2×3)×(5)×(2×2)×(3)×2× 7!

10! = (3×2×1) × (5×4×3×2×1) × 7!

10! = 3! × 5! × 7! = 7!5!3! (Verified)

c) 16!= 14!5!2!

16! = 16×15×14!

16! = (2×2×2×2) × (5×3) × 14!

Rearranging the factors.

16! = (2) × (5) × (2×2) × (3) × (2) ×14!

16! = (2!) ×(5×4×3×2) × 14!

16! = 2! × 5! × 14!

16! = 14!5!2! (Verified)

d) 9!=7!3!3!2

9! = 9×8×7!

9! = (3×3) × (2×2×2) × 7!

Rearranging the factors

9! = (3×2) ×(3×2) ×(2) × 7!

9! = 3! × 3! ×2! × 7!

9! = 7!3!3!2! (Verified)

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15,800 if the equation is (3028+200-54)5
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