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3 years ago

What is HB? Triangle Mid-segment Theorem

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

HB = 13

Step-by-step explanation:

In triangle GHJ, points A, B, C are mid points of the sides GH, HJ, GJ respectively.

Therefore by Mid-segment Theorem.

$AC = \frac{1}{2} HJ \\ \\ 3y - 5 = \frac{1}{2} (4y + 2) \\ \\ 3y - 5 = \frac{1}{2} \times 2 (2y + 1) \\ \\ 3y - 5 = 2y + 1\\ \\ 3y - 2y = 5 + 1 \\ \\ y = 6 \\ \\ HJ = 4y + 2 \\ \\ HJ = 4 \times 6 + 2 \\ \\ HJ = 24 + 2 \\ \\ HJ = 26 \\ \\ HB= \frac{1}{2} HJ \\ \\ HB= \frac{1}{2} \times 26 \\ \\ \huge \red{ \boxed{ HB= 13}}$

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4 years ago

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Answer:

—————

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Step-by-step explanation:

Step by Step Solution:

More Icon

STEP

Equation at the end of step 1

((12•(n3))-(24•(n2))) (12n-42)

—————————————————————•———————————————————

(((4•(n2))-22n)+28) ((6•(n3))+(24•3n2))

STEP

Equation at the end of step

((12•(n3))-(24•(n2))) (12n-42)

—————————————————————•——————————————————

(((4•(n2))-22n)+28) ((2•3n3)+(24•3n2))

STEP

12n - 42

Simplify ——————————

6n3 + 48n2

STEP

Pulling out like terms

4.1 Pull out like factors :

12n - 42 = 6 • (2n - 7)

STEP

Pulling out like terms

5.1 Pull out like factors :

6n3 + 48n2 = 6n2 • (n + 8)

Equation at the end of step

((12•(n3))-(24•(n2))) (2n-7)

—————————————————————•————————

(((4•(n2))-22n)+28) n2•(n+8)

STEP

Equation at the end of step

((12•(n3))-(24•(n2))) (2n-7)

—————————————————————•————————

((22n2-22n)+28) n2•(n+8)

STEP

Equation at the end of step

((12•(n3))-(23•3n2)) (2n-7)

————————————————————•————————

(4n2-22n+28) n2•(n+8)

STEP

Equation at the end of step

((22•3n3) - (23•3n2)) (2n - 7)

————————————————————— • ————————————

(4n2 - 22n + 28) n2 • (n + 8)

STEP

12n3 - 24n2

Simplify ——————————————

4n2 - 22n + 28

STEP

Pulling out like terms

10.1 Pull out like factors :

12n3 - 24n2 = 12n2 • (n - 2)

STEP

Pulling out like terms

11.1 Pull out like factors :

4n2 - 22n + 28 = 2 • (2n2 - 11n + 14)

Trying to factor by splitting the middle term

11.2 Factoring 2n2 - 11n + 14

The first term is, 2n2 its coefficient is 2 .

The middle term is, -11n its coefficient is -11 .

The last term, "the constant", is +14

Step-1 : Multiply the coefficient of the first term by the constant 2 • 14 = 28

Step-2 : Find two factors of 28 whose sum equals the coefficient of the middle term, which is -11 .

-28 + -1 = -29

-14 + -2 = -16

-7 + -4 = -11 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -7 and -4

2n2 - 7n - 4n - 14

Step-4 : Add up the first 2 terms, pulling out like factors :

n • (2n-7)

Add up the last 2 terms, pulling out common factors :

2 • (2n-7)

Step-5 : Add up the four terms of step 4 :

(n-2) • (2n-7)

Which is the desired factorization

Canceling Out :

11.3 Cancel out (n-2) which appears on both sides of the fraction line.

Equation at the end of step

6n2 (2n - 7)

—————— • ————————————

2n - 7 n2 • (n + 8)

STEP

Canceling Out

12.1 Cancel out (2n-7) which appears on both sides of the fraction line.

Canceling Out :

12.2 Canceling out n2 as it appears on both sides of the fraction line

Final result :

—————

n + 8

8 0

4 years ago