All categories

4 years ago

The center of the circle is O. Find the measure of angle BDE *

Mathematics

2 answers:

lorasvet [3.4K]4 years ago

7 0

Answer:

90 (angle in a semicircle)

Aleksandr [31]4 years ago

7 0

Good morning ☕️

Answer:

m∠BDE = 90°

Step-by-step explanation:

We notice that the segment BE is a diameter of the circle

and D is a point of the circle different from points B and E

therefore m∠BDE = 90°

__________________

You might be interested in

What is the mode(s) of this data?

Neko [114]

Answer:

58,65

Step-by-step explanation:

Mode= the most common number

7 0

3 years ago

jeff wants to buy a phone card for long distance calls. he can buy a 200-minute card for $10.00 or a 300-minute card for $12.00.

Alenkasestr [34]

200 min / $10.00 = 20 minutes / dollar

300 min / $12.00 = 25 minutes / dollar 

The cost of the 300 minute card is cheaper per minute and a better value. 

If you don't expect to use more than 200 minutes then save your money and buy the smaller card.

3 0

4 years ago

Read 2 more answers

Is the additive inverse of an integer always negative? Explain.

andre [41]

Answer:

Step-by-step explanation:

We do not consider zero to be a positive or negative number. For each positive integer, there is a negative integer, and these integers are called opposites. For example, -3 is the opposite of 3, -21 is the opposite of 21, and 8 is the opposite of -8. ... If an integer is less than zero, we say that its sign is negative

3 0

3 years ago

Hat is the 68th term of sequence -8.6,-7.2,-5.8,-4.4,

Umnica [9.8K]

First find the law
8.6-1.4n
so, 64th is 8.6-1.4x64
I think is -81

5 0

3 years ago

PLEASE ANSWER + BRAINLIEST!! Factor completely. 4k - 20k^9 = 3b^2 - 108 =

Neporo4naja [7]

$4k-20k^9=4k(1-5k^8)=4k\left[1^2-(k^4\sqrt5)^2\right]\\\\=4k(1-k^4\sqrt5)(1+k^4\sqrt5)=4k\left[1^2-(k^2\sqrt[4]5)^2\right](1+k^4\sqrt5)\\\\=4k(1-k^2\sqrt[4]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k\left[1^2-(k\sqrt[8]5)^2\right](1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k(1-k\sqrt[8]5)(1+k\sqrt[8]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)$

$3b^2-108=3(b^2-36)=3(b^2-6^2)=3(b-6)(b+6)$

$Used:\ (a-b)(a+b)=a^2-b^2$

7 0

3 years ago