NEED HELP PLZ Each edge of the cube measures 4 inches in length
What is the distance, in inches, from vertex B to vertex H? Round your answer to the nearest tenth of an inch

The distance from vertex B to vertex H is 6.9 inches.

Step-by-step explanation:

The length of the line BH can be thought of as the hypotenuse of a triangle rectangle where the catheti are the lines HD (whit a measure of 4in) and line DB.

The length of the line DB can be thought of as the hypotenuse of a triangle rectangle where the catheti are lines DA and AB (both are 4in long)

Then if we use the Pythagorean's theorem, the length of line DB is:

(DB)^2 = (DA)^2 + (AB)^2

(DB)^2 = (4in)^2 + (4in)^2 = 32in^2

(DB) = √(32in^2) = 5.66 in

Whit this, we can find the length of line HB as:

(HB)^2 = (HD)^2 + (DB)^2

(HB)^2 = (4in)^2 + (DB)^2 = 16in^2 + 32in^2 = 48in^2

HB = √(48in^2) = 6.93 in

If we round to the nearest thent, we get:

HB = 6.9 in

The distance from vertex B to vertex H is 6.9 inches.

x = -7,   y = 5

Step-by-step explanation:

a) x + 9y = 38,   b) 3x + 7y = 14

From a)

x = 38 - 9y (Equation 1)

From b)

{3*(38 - 9y)} + 7y = 14 (From equation 1)

114 - 27y + 7y -14 = 0

100 - 20y = 0

20y = 100

y = 5

From a)

x + (9*5) = 38  (substituting the value of y = 5)

x + 45 = 38

x = 38 - 45

x = -7

Hope, this will help