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Pachacha [2.7K]
3 years ago
13

Point H is on line segment \overline{GI} GI . Given GI=3x,GI=3x, HI=2,HI=2, and GH=5x-6,GH=5x−6, determine the numerical length

of \overline{GH}. GH .
Mathematics
2 answers:
AlekseyPX3 years ago
8 0

Answer:

GH=4

Step-by-step explanation:

Hope this hedlps anyone :) Have a great day!

enot [183]3 years ago
3 0

Answer:

Step-by-step explanation:

3x + 2 = 5x - 6

-2x + 2 = -6

-2x = -8

x = 4

3(4) + 2 = 12 + 2 = 14

5(4) - 6= 20 - 6 = 14 for GH

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Jane's salary is increased by 2% to £39474. how much is her salary before the rise?
iVinArrow [24]
Let her salary before the rise = x pounds then
x + 0.02x = 39474
1.02x = 39474
x = 39474 / 1.02 =  £38,700  answer
4 0
3 years ago
PLEASE HELP WILL MARK BRAINLEIST
horsena [70]

Answer:

The picture shows the answer on the graph

Step-by-step explanation:

-2=-3/4×(x-6)

Distribute -3/4 through the parenthesis

-2= -3/4x+9/2

Multiply both sides of the equation by 4

-8=-3x+18

Move the variable to the left side and change its sign

3x-8=18

Move constant to the right side and change its sign

3x=18+8

Add the numbers

3x=26

Divide both sides of the equation by 3

x=26/3

Alternative Form- x=8 2/3 or x=8.6

7 0
3 years ago
Fine length of BC on the following photo.
MrMuchimi

Answer:

BC=4\sqrt{5}\ units

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

AC^2=AD^2+DC^2

substitute the given values

AC^2=16^2+8^2

AC^2=320

AC=\sqrt{320}\ units

simplify

AC=8\sqrt{5}\ units

step 2

In the right triangle ACD

Find the cosine of angle CAD

cos(\angle CAD)=\frac{AD}{AC}

substitute the given values

cos(\angle CAD)=\frac{16}{8\sqrt{5}}

cos(\angle CAD)=\frac{2}{\sqrt{5}} ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

cos(\angle BAC)=\frac{AC}{AB}

substitute the given values

cos(\angle BAC)=\frac{8\sqrt{5}}{16+x} ----> equation B

step 4

Find the value of x

In this problem

\angle CAD=\angle BAC ----> is the same angle

so

equate equation A and equation B

\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}

solve for x

Multiply in cross

(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units

DB=4\ units

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

BC^2=DC^2+DB^2

substitute the given values

BC^2=8^2+4^2

BC^2=80

BC=\sqrt{80}\ units

simplify

BC=4\sqrt{5}\ units

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Step-by-step explanation:

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