name a transformation that does not always produce an image that would be congruent to the original figure

1 answer:

3 0

A dilation is the best answer. A dilation shrinks or expands the shape on a plane. But now, that shape is the same shape, but a different size.
Congratulations! You have discovered simillar figures. They are the same shape, but different sizes, so they are not congruent.
Hope this helps

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Solve this x²+8/10 -x = 10/2

AysviL [449]

$x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\


$
$x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\$

7 0

3 years ago

Round to the nearest hundredth (Geometric Series problem)

WITCHER [35]

Answer:

15.98

Step-by-step explanation:

∑₁¹⁰ $8(\frac{1}{4} )^{n-1}$

$=8[1+\frac{1}{4} +\frac{1}{16} + ........ +t_{10}]$

We have to use the formula of sum of a G.P. series with common ratio r < 1.

If the first term is a, the common ratio is r and the number of terms is n then the sum [a + ar + ar² + ar³ + ......... up to n terms] is given by

$[a \times\frac{1-r^{n} }{1-r}]$