All categories

2 years ago

What common angle do ACE and ACD share?

Mathematics

2 answers:

SCORPION-xisa [38]2 years ago

6 0

The answer is the second option, <ACE = <ECA.

This is because C is in the middle of both, so it's referring to the same angle. The other options are all comparing different angles, that aren't congruent, so they're incorrect.

never [62]2 years ago

3 0

Answer:

∠ACE≅∠ECA

Step-by-step explanation:

If we take ΔACE and ΔACD,

∠ACE and ∠ECA are the common angle as compared to the rest one given.

If we take, ∠AFB and ∠EFD, both the angles are not contained in the given triangle.

Similarly, if we take ∠EBC and∠ECB, the ∠EBC is not included in the triangle ACE.

In, ∠CAD and ∠CEB, both the angles cannot be equal.

You might be interested in

A company produces metal slabs for industrial use. the slabs come in 3 thicknesses: 5in, 10in, and 15in. the weights of each thi

Len [333]

Given slabs thicknesses 5in, 10in, and 15in.

Therefore, ratio of thicknesses in simplest form would be 1:2:3.

Metal 1 thicknesses = 150lbs, 300lbs, 600lbs

Dividing by 150, we get ratio of Metal 1 thicknesses = 1:2:4.

Metal 2 thicknesses = 130lbs, 260lbs, 390lbs.

Dividing by 130, we get ratio of Metal 2 thicknesses = 1:2:3.

Metal 3 thicknesses = 310lbs, 620lbs, 930lbs.

Dividing by 310, we get ratio of Metal 3 thicknesses = 1:2:3.

Metal 4 thicknesses = 140lbs, 280lbs, 420lbs.

Dividing by 140, we get ratio of Metal 4 thicknesses = 1:2:3.

We can see the ratios of thicknesses of Metal 1 are 1:2:4 but it should be 1:2:3.

Therefore, Metal 1 weight is not proportional to thickness.

5 0

2 years ago

Read 2 more answers

Greta is taking a road trip from Cincinnati to San Francisco. On the first day, she drove 650 miles in 10 hours. On the second d

abruzzese [7]

Given:

On the first day, she drove 650 miles in 10 hours.

On the second day, she got a later start and drove 540 miles in 8 hours.

To find:

Difference between average speed of second day and first day.

Solution:

We know that,

$Speed=\dfrac{Distance}{Time}$

On the first day, she drove 650 miles in 10 hours. So, the average speed is

$Speed=\dfrac{650}{10}$

$Speed=65$

So, the average speed on first day is 65 miles per hour.

On the second day, she got a later start and drove 540 miles in 8 hours.

$Speed=\dfrac{540}{8}$

$Speed=67.5$

So, the average speed on second day is 67.5 miles per hour.

Difference between average speed is

$67.5-65=2.5$

Therefore, the average speed on the second day is 2.5 miles per hour is faster than first day.

3 0

1 year ago

A marina is in the shape of a coordinate grid. Boat A is docked at (2.4, −3) and Boat B is docked at (−3.4, −3). The boats are _

Tasya [4]

They are 5.8 units apart

5 0

2 years ago

Read 2 more answers

A certain car traveling at 34.0 mph skids to a stop in 29 meters from the point where the brakes were applied. in approximately

Orlov [11]

Using the kinematic equations: (final velocity)^2 = (initial velocity)^2 - 2 * acceleration * distance Assuming the acceleration/deceleration on the car is constant from a constant force on the brakes. Converting from mph to m/s using 0.447 (so 34 mph is 15.2 m/s) (0)^2 = (15.2)^2 - 2 * acceleration * 29 acceleration = 4.0 m/s^2 Had the car been going 105.4 mph (47.1 m/s) (0)^2 = (47.1)^2 - 2 * 4 * distance distance = 277 meters

6 0

2 years ago

A research organization wanted to estimate the average number of hours a college student sleeps per night during the school year

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

Given given that

Lower confidence interval = 7.1

Upper confidence interval = 7.5

As such, We are 95% confident that it's somewhere between 7.1 and 7.5 hours per night

Average value = LCI + UCI /2 = 7.1 + 7.5 / 2

= 7.3hours per night, so we are 95% confident that it's somewhere between 7.1 and 7.5 hours per night which is eventually 7.3hrs per night.

4 0

2 years ago