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1 year ago

In the given figure, O is centre of the circle. Find the value of x and y?

Mathematics

1 answer:

Effectus [21]1 year ago

4 0

Answer:

The value of x = 13cm and y = 12cm.

Step-by-step explanation:

P is the exterior point to the circle with centre O

Radius = 5cm

Length of the tangent = 12 cm

Distance between the point and centre of the circle = x cm

We know that

The angle between the radius of the circle and the tangent at the point of contact is 90°

By Pythagoras Theorem

Hypotenuse² = Side²+Side²

⇛x² = 12²+5²

⇛x² = 144+25

⇛x² = 169

⇛x² = 13²

We know that

The lengths of tangents drawn from the exterior point to the circle are equal.

<u>Answer</u><u>:</u> The value of x=13cm and y = 12cm.

<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> in the given figure, O and O' are centre of two circles and the circles intersect each other at points B and c. If AOCD is a straight line and /_ AOB =110, find...

brainly.com/question/957506?referrer

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4p-10<6p-8 or 10p+16<9p-3

Finger [1]

Answer:

p>-1 or p<-19

Step-by-step explanation:

4p-10<6p-8 (add 10 to each side)

4p< 6p+2 (subtract 6p from each side)

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2 years ago

In parallelogram ABCD , BE=7x−2 and ED=x2−10 .

Stells [14]

I'm assuming that the diagonals intersect at E. Since diagonals of a parallelogram bisect each other, BE=ED.

7x-2=x²-10
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As distance must be positive, we reject the negative case, so x=8.

Thus, BE=ED=54.

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1 year ago

The factorization of x² + 3x -4 is modeled with algebra

nydimaria [60]

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1 year ago

1. What is the solution of n² – 49 = 0? (1 point)

maks197457 [2]

Answer:

1. $n=\pm7$

2. $x=\pm7i[tex]3.[tex]\pm 12x =a$

4. 22

Step-by-step explanation:

1. $n^2 - 49 = 0$

$n^2 =49$

$n= \sqrt{49}$

$n=\pm7$

2. $x^2+64 = 0$

$x^2 =-64$

$x= \sqrt{49}$

$x=\pm7i[tex]3. Area of square = [tex]a^{2}$

Where a is the side

We are given an area o square = $144x^{2}$

So, $144x^{2}=a^{2}$

$\sqrt{144x^{2}}=a$

$\pm 12x =a$

4. We are given that the solution of quadratic equation -9 and 9

So, equation becomes:

$(x+9)(x-9)=0$

$x^2- 81=0$

So, in the given equation $x^2 + z = 103$

z should be the number from which if we subtract 103 so we get 81

Substitute z = 22

$x^2 + 22 = 103$

$x^2 + 22 -103=0$

$x^2 -81=0$

Thus the value of z is 22

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3 years ago

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2 years ago

In the given figure, O is centre of the circle. Find the value of x and y? ​

In the given figure, O is centre of the circle. Find the value of x and y?