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3 years ago

Which transformation produces an image that is not congruent to the original figure?

Mathematics

2 answers:

Digiron [165]3 years ago

7 0

The answer is dilation because it changes the size of the original figure, so it is no longer congruent.

swat323 years ago

7 0

A dilation produces an image not congruent to the original figure because the size of the preimage is different from the image, it is not a rigid transformation

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What is the length of side x in the above right triangle

nataly862011 [7]

Answer:

its answer will be 9.first by using phythagorus theorem h^2=p^2+b^2'then h=15,p=12and b=x. again by putting the values of h ,p and b we get ,(15)^2=(12)^2+x^2. 225=144+x^2, x^2=225-144, x^2=81,x=√81,x=9 answer

7 0

2 years ago

How can i prove this property to be true for all values of n, using mathematical induction.

chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

$\implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6}$

$\implies{ \frac{1(2)(3)}{6} }$

$\implies{ 1}$

Therefore, it is true for the first part.

In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

and we will prove that,

$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

$\: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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8 0

1 year ago

Write the equation of the parabola in vertex form. Vertex (2,1), point (3, -4)

Ket [755]

Answer:

U=1-5(x-2)^2

Step-by-step explanation:

The equation of the parabola with vertex (h,k) is y=a(−h+x)2+k

Thus, the equation of the parabola is y=a(x−2)2+1

To find a, use the fact that the parabola passes through the point (3,−4): −4=a+1

Solving this equation, we get that a=−5

Thus, the equation of the parabola is y=1−5(x−2)^2

6 0

3 years ago

Please Help! 20 Points!

iVinArrow [24]

$\text {x - coordinate = } \bigg( 5\dfrac{1}{2} + 3\dfrac{3}{4} \bigg) \div 2$

$\text {x - coordinate = } \bigg( 5\dfrac{2}{4} + 3\dfrac{3}{4} \bigg) \div 2$

$\text {x - coordinate = } 8\dfrac{5}{4} \div 2$

$\text {x - coordinate = }\dfrac{37}{4} \times \dfrac{1}{2}$

$\text {x - coordinate = }\dfrac{37}{8}$

$\text {x - coordinate = } 4\dfrac{5}{8}$

$\text {y - coordinate = } \bigg( -4\dfrac{1}{4} - 1\dfrac{1}{4} \bigg) \div 2$

$\text {y - coordinate = } -5\dfrac{2}{4} \div 2$

$\text {y - coordinate = } -\dfrac{22}{4} \times \dfrac{1}{2}$

$\text {y - coordinate = } -\dfrac{11}{4}}$

$\text {y - coordinate = } -2\dfrac{3}{4}}$

$\text {The coordinate is }\bigg( 4\dfrac{5}{8}, -2\dfrac{3}{4}} \bigg)$

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2 years ago

What is the value 49,254

erastova [34]

40,000
9,000
200
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4

I'm guessing that's what you were looking for?

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2 years ago