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Kobotan [32]
3 years ago
6

Solve this problem. What is x?

Mathematics
1 answer:
DedPeter [7]3 years ago
5 0

a^2 + b^2 = c^2

a^2 = 6^2 + √117^2

a^2 + 36 = 117

a^2 = 117 - 36

a^2 = 81

a = √81

a = 9

x = 9

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Someone pls help and show ur work
ale4655 [162]

Answer:

8x8=16

easy

https://youtu.be/WYNfQabLpuk

6 0
3 years ago
Select the positive or negative integer for the situation.
prisoha [69]

Answer:

1500

Step-by-step explanation:

deposit means put in and withdraw means take

6 0
3 years ago
Read 2 more answers
Point D is located on MN at (2,43). What ratio relates MD to DN?
Eva8 [605]

The ratio of MD to DN is equal to 2.

<h3>How to find the partition ratio for a line segment</h3>

In accordance with the image set aside, the locations of the points M and N are M(x, y) = (- 6, - 4) and N(x, y) = (6, 4), respectively. Now we determine the vectors associated to line segments MD and DN by vector sum:

<u>MD</u> = D(x, y) - M(x, y)

<u>MD</u> = (2, 4 / 3) - (- 6, - 4)

<u>MD</u> = (8, 16 / 3)

<u>DN</u> = N(x, y) - D(x, y)

<u>DN</u> = (6, 4) - (2, 4 / 3)

<u>DN</u> = (4, 8 / 3)

Lastly, we find the length of each line segment by Pythagorean theorem:

MD = √[8² + (16 / 3)²]

MD = (8 / 3)√13

DN = √[4² + (8 / 3)²]

DN = (4 / 3)√13

And the ratio of MD to DN is:

MD / DN = [(8 / 3)√13] / [(4 / 3)√13]

MD / DN = 2

The ratio of MD to DN is equal to 2.

<h3>Remark</h3>

The statement presents typing mistakes, we kindly present the correct form below:

<em>Point D is located on line segment MN at (2, 4 / 3). What ratio relates MD to DN?</em>

To learn more on line ratios: brainly.com/question/3148758

#SPJ1

4 0
1 year ago
what side lengths should be used to model a rectangle with an area of x^2+3x-10 square units? x-5 and x+2 x and x+3 x+3 and x-10
klemol [59]

Answer:

Answer: x-2 and x+5 side lengths should be used to model a rectangle.

Step-by-step explanation:

can u make me brainiest pls

5 0
2 years ago
Read 2 more answers
MN is a diameter of a circle with centre ' O ' . If BD = CD , prove that ∠OAD = ∠OCD
Varvara68 [4.7K]

Answer:

\sf \: Proved \: \angle \: OAD  \:  =  \angle \: OCD

Step-by-step explanation:

<em>Given:</em>

Mn is diameter of circle having centre O

and BD = OD,

<em><u>To prove that:</u></em>

<u>\angle \: OAD  \:  =  \angle \: OCD</u>

<em>Solution:</em>

Join the points O and B and draw OB,

On joining the line,

in ∆OCD and ∆OBD,

OC =OB → (Radius of same circle)

BD =CD → (from given)

OD =OD → (Common side in both the triangles)

Hence ∆OCD and ∆OBD are congruent from SSS property.

so we can say that,

\angle \: OBD  \:  =  \angle \: OCD

Consider above prove as statement A

Corresponding angles of congruent traingle.

in ∆ OAB,

OA = OB (radius of same circle)

hence ∆OAB is an isosceles traingle.

We know that opposite angle of isosceles traingle are always equal. hence,

\angle \: OBD  \:  =  \angle \: OAB \\ \angle \: OAB  \:  =  \angle \: OAD (same \: angles) \\ \angle \: OBD  \:  =  \angle \: OAD

Consider above prove as statement B

From Statement A & B we can say that

\angle \: OAD  \:  =  \angle \: OCD

<em>Thanks for joining brainly community!</em>

5 0
2 years ago
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