How does changing the scale change the distance represented?

Geometry help please how do I do these?

trapecia [35]

Answer:

Question 7:

∠L = 124°

∠M = 124°

∠J = 118°

Question 8:

∠Q = 98°

∠T = 98°

∠R = 82°

Question 15:

m∠G = 110°

Question 16:

∠G = 60°

Question 17:

∠G = 80°

Question 18:

∠G = 70°

Step-by-step explanation:

The angles can be solving using Symmetry.

Question 7.

The sum of interior angles in an isosceles trapezoid is 360°, and because it is an isosceles trapezoid

∠K = ∠J = 118°

∠L = ∠M

∠K+∠J+∠L +∠M = 360°

236° + 2 ∠L = 360°

Therefore,

∠L = 124°

∠M = 124°

∠J = 118°

Question 8.

In a similar fashion,

∠Q+∠T+∠S +∠R = 360°

and

∠R = ∠S = 82°

∠Q = ∠T

∠Q+∠T + 164° = 360°

2∠Q + 164° = 360°

2∠Q = 196°

∠Q = ∠T =98°.

Therefore,

∠Q = 98°

∠T = 98°

∠R = 82°

Question 15.

The sum of interior angles of a kite is 360°.

∠E + ∠G + ∠H + ∠F = 360°

Because the kite is symmetrical

∠E = ∠G.

And since all the angles sum to 360°, we have

∠E +∠G + 100° +40° = 360°

2∠E = 140° = 360°

∠E = 110° = ∠G.

Therefore,

m∠G = 110°

Question 16.

The angles

∠E = ∠G,

and since all the interior angles sum to 360°,

∠E + ∠G + ∠F +∠H = 360°

∠E + ∠G + 150 + 90 = 360°

∠E + ∠G = 120 °

∠E = 60° = ∠G

therefore,

∠G = 60°

Question 17.

The shape is a kite; therefore,

∠H = ∠F = 110°

and

∠H + ∠F + ∠E +∠G = 360°

220° + 60° + ∠G = 360°,

therefore,

∠G = 80°

Question 18.

The shape is a kite; therefore,

∠F = ∠H = 90°

and

∠F +∠H + ∠E + ∠G = 360°

180° + 110° + ∠G = 360°

therefore,

∠G = 70°.

3 0

3 years ago

The longer leg of a right triangle is 4cm longer than the shorter leg. the hypotenuse is 8cm longer than the shorter leg. find t

Alexeev081 [22]

Longer leg = x + 4

Shorter leg = x

Hypotenuse = x + 8

Using Pythagorean Theorem:

a^2 + b^2 = c^2

(x + 4)^2 + x^2 = (x + 8)^2

x^2 + 8x + 16 + x^2 = x^2 + 16x + 64

2x^2 + 8x + 16 = x^2 + 16x + 64

2x^2 +8x = x^2 + 16x + 48

2x^2 - 8x = x^2 + 48

x^2 - 8x = 48

x^2 - 8x - 48 = 0

You can complete the square from here or use the quadratic formula.

Completing the square:

x^2 - 8x = 48

x^2 - 8x + (-8/2)^2 = 48 + (-8/2)^2

x^2 - 8x + 16 = 48 + 16

(x - 4)(x - 4) = 64 or (x - 4)^2 = 64

x - 4 = +√64 OR x - 4 = -√64

x - 4 = +8 OR x - 4 = -8

x = 12 OR x = -4

However, you can't use negative 4 as a length because your length needs to be a positive.

So x will be 12.

Shorter leg: 12

Longer leg: 12 + 4

Hypotenuse: 12 + 8

8 0

3 years ago

Help pleaseeeeeeeee

Doss [256]

Answer:9

Step-by-step explanation: im not really sure I kinda just guess and check, so like, 2/3 times 3 is 6/3 which is 2. so 6/2 is 3. 3 times 3 is nine.

4 0

3 years ago

Plz help simplify 8/10

Free_Kalibri [48]

$\frac{8}{10}$ → $\frac{4}{5}$

8 and 10 can both be divided by 2. So, 8 ÷ 2 is 4 and 10 ÷ 2 is 5.

Hope this helps :)

3 0

3 years ago

Read 2 more answers

A pilot is flying a CRJ-900 from Phoenix, Arizona, to Louisville, Kentucky. The aircraft cannot take off or land if the temperat

ioda

We need to find an inequality that represents the possible landing or take of conditions of an aircraft, the required graph and one case where the inequality will be tested.

The required inequality is $-40$ .

The graph represents the allowable temperature range in which the aircraft can takeoff or land.

The pilot could not have taken off on June 1990, the temperature in Phoenix, Arizona.

Let the temperature be denoted by $x$

The temperature cannot be at or below $-40^{\circ}\text{F}$ , so

$x>-40$

The temperature cannot be at or above $118^{\circ}\text{F}$ , so

$x$

The inequality will be $-40$ .

In the graph it can be seen that all points between -40 and 118 falls in the shaded region but it will exclude the points -40 and 118.

The graph represents the allowable temperature range in which the aircraft can takeoff or land.

The temperature in June 1990 in Phoenix, Arizona was $122^{\circ}$ which is more than $118^{\circ}\text{F}$ .

Hence, the pilot could not have taken off on this day.

Learn more:

brainly.com/question/20383699

brainly.com/question/11612965

4 0

2 years ago