All categories

3 years ago

In the diagram below, DE is parallel to XY. What is the value of x?

Mathematics

1 answer:

iren2701 [21]3 years ago

7 0

Answer:

A. 85

Step-by-step explanation:

Since DE and XY are parallel, this means that the given angle and angle x are alternate Angles. According to the law of alternate Angles, this means that they are equal. It also makes sense as the angle x is the same angle as the angle directly opposite angle 85. This is because they are on the same line across parallel lines. This also means they are equal because vertically opposite angles are always equal.

Therefore x =85

Hope this helped!

You might be interested in

Pleaseee help guys:(

nadya68 [22]

≈0.524094
If you ever needed help with math you can download Photomath too that’s what I use

8 0

4 years ago

Read 2 more answers

Help with math please?

Arlecino [84]

Ok So you Look at A Graph . You look at the last number and thats your starting Number on the Y access. You go up 5 blocks and plot a Point. Then you look at the -3x Thats a Negative number going down so your start from the 5 then go down 3 then to the right 1.

Hope this Helps

-Dante :D

7 0

4 years ago

Read 2 more answers

Amtrak trains from Chicago to Portland arrived at Havre, MT, where they make a crew change, the following numbers of minutes lat

arsen [322]

Answer:

45.6 min

Step-by-step explanation:

(26+17+245+19+38+42+11+0+14+44) / 10 = 456 / 10 = 45.6

6 0

3 years ago

Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=

jasenka [17]

Answer:

The function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

$\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

The function is continuous at point x=36 if:

The function $\\ f(x)$ exists at x=36.
The limit on both sides of 36 exists.
The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

$\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}$

$\\ f(36) = \frac{36*6}{900} = \frac{6}{25}$

The limit of the $\\ f(x)$ is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

$\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Since

$\\ f(36) = \frac{6}{25}$

And

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Then, the function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

8 0

3 years ago

ANSWER THE QUESTION FOR BRAINLIESTANSWER THE QUESTION FOR BRAINLIEST

rewona [7]

Answer:

A) Ed

Explanation: How i got it was by first looking at what Eric’s weight loss program, I then guessed he lost about the same amount of weight each week (1.5), i then did that times 4 since that’s how many weeks were recorded on the chart which gave me 6. I then looked at the chart again and saw that Ed first started with the weight 205 and by week 4 his weight was 197, I then did 205-197 which equals 8, I then decided that Ed lost more weight.

3 0

3 years ago

Read 2 more answers