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zhenek [66]
2 years ago
9

Please help with this asappppppppll

Mathematics
1 answer:
Alenkinab [10]2 years ago
7 0

We are given with a circle and we need to find the <em>equation of the circle</em> , but first let's recall that , the equation of a circle with radius<em> 'r'</em> and centre at <em>(h,k) </em>is given by {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}}

Now , here as as the circle cuts the +ve x-axis at (9,0) . So , it's radius is 9 units or the 2nd way is to measure the distance from centre of the circle to the point where the circle cuts the graph , as the centre is at Origin , so here <em>(h,k) = (0,0)</em> .Which means that the centre is located at the point whose coordinates are<em> (0,0)</em> which is also known as origin . Now , finding the equation of the circle :-

{:\implies \quad \sf (x-0)^{2}+(y-0)^{2}=(9)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{x^{2}+y^{2}=81}}}

<em>This is the required equation of Circle</em>

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Find the greatest common factor of these three expressions. 5v2 , 14v4 , and 35
a_sh-v [17]

Question

Find the greatest common factor of these three expressions. 5v2 , 14v4 , and 35


Answer:

<h2>1</h2>

Step-by-step explanation:

assuming that 5v2 is 5², and 14v4 is 14^4

5² = 25

14^4 = 38416

35 = 35

The factors of 25 are: 1, 5, 25

The factors of 35 are: 1, 5, 7, 35

The factors of 38416 are: 1, 2, 4, 7, 8, 14, 16, 28, 49, 56, 98, 112, 196, 343, 392, 686, 784, 1372, 2401, 2744, 4802, 5488, 9604, 19208, 38416

--------------

the greatest common factor is 1



6 0
2 years ago
Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse
Drupady [299]

Answer:

a'(d) = \frac{d}{5} + \frac{3}{5}

a(a'(d)) = a'(a(d)) = d

Step-by-step explanation:

Given

a(d) = 5d - 3

Solving (a): Write as inverse function

a(d) = 5d - 3

Represent a(d) as y

y = 5d - 3

Swap positions of d and y

d = 5y - 3

Make y the subject

5y = d + 3

y = \frac{d}{5} + \frac{3}{5}

Replace y with a'(d)

a'(d) = \frac{d}{5} + \frac{3}{5}

Prove that a(d) and a'(d) are inverse functions

a'(d) = \frac{d}{5} + \frac{3}{5} and a(d) = 5d - 3

To do this, we prove that:

a(a'(d)) = a'(a(d)) = d

Solving for a(a'(d))

a(a'(d))  = a(\frac{d}{5} + \frac{3}{5})

Substitute \frac{d}{5} + \frac{3}{5} for d in  a(d) = 5d - 3

a(a'(d))  = 5(\frac{d}{5} + \frac{3}{5}) - 3

a(a'(d))  = \frac{5d}{5} + \frac{15}{5} - 3

a(a'(d))  = d + 3 - 3

a(a'(d))  = d

Solving for: a'(a(d))

a'(a(d)) = a'(5d - 3)

Substitute 5d - 3 for d in a'(d) = \frac{d}{5} + \frac{3}{5}

a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}

Add fractions

a'(a(d)) = \frac{5d - 3+3}{5}

a'(a(d)) = \frac{5d}{5}

a'(a(d)) = d

Hence:

a(a'(d)) = a'(a(d)) = d

7 0
2 years ago
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Answer:

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Consider the original complex figure and the reduction.
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Answer: The answer is C. 5

Step-by-step explanation:

Just took the test on edg.

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Answer:

y=1/2x

Step-by-step explanation:

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