What is HB? Triangle Mid-segment Theorem

Mathematics

1 answer:

Scorpion4ik [409]2 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.

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

You might be interested in

If p(x)=2x^3+cx^2+16x+84 and x+2 is a factor of p(x),find the value of c.

Nuetrik [128]

Answer:

c = -9

Step-by-step explanation:

Knowing that -2 is a root, you can do the synthetic division with an unknown value for c. Then solve for the value of c that makes the remainder be zero. See the attachment for the work.

The second attachment shows that (x+2) is a factor of the resulting cubic.

4 0

2 years ago

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating

MakcuM [25]

Answer:

A=8.4063 $u^{2}$

Step-by-step explanation:

Be the functions:

$y=\frac{3}{x};y=\frac{3}{x^{2}}:x=7$

according the graph:

$\int\limits^1_7 {\frac{3}{x} } \, dx -\int\limits^1_7 {\frac{3}{x^{2} } } \, dx =3\int\limits^1_7 {\frac{1}{x} } \, dx -3\int\limits^1_7 {\frac{1}{x^{2} } } \, dx=3(\int\limits^1_7 {\frac{1}{x} } \, dx -\int\limits^1_7 {\frac{1}{x^{2} } } \, dx)=3[lnx-\frac{1}{x}](1-7)=3[(ln7-ln1)-(\frac{1}{7}-1)]=3[(1.945-0)-(0.1428-1)]=3*(1.945+0.8571)=3*2.8021=8.4063u^{2}$