AB is a chord of a circle of radius 5cm the distance of AB from the centre is 3cm then AB is?

8 cm

Step-by-step explanation:

We first want to draw a diagram with the given information.

I am unsure if the image shows below, or if it is an attachment.

We then see we have a right triangle. What is a theorem that is commonly associated with right triangles?

So we use the pythagorean theorem to find the missing leg of the triangle has a length of 4 cm. But how does this help us? Well, when finding the distance from the center of a circle to a chord, we always find the line that is perpendicular to the chord and divides it in half (and passes through the center), and find the length.

So, the leg of length 4 we just found? That's half of our answer.

Since 4 × 2 = 8, we know that the length of AB is 8 cm.

answer: this is not a question

step-by-step explanation: this is a answer

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.