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4 years ago

14

E^(2lnx)=4log10 Solve for x

1 answer:

Licemer1 [7]4 years ago

3 0

X = in4/2in

I hope this helps.

Have a great day. :)

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Describe the composite transformation that has occurred.

Sidana [21]

Answer:

rotate CCW 90°, reflect across the x-axis
(x, y) ⇒ (-y, -x) . . . . . both transformations together

Step-by-step explanation:

The vertex order ABC is clockwise in the original figure and also in the first image: A'B'C'. The altitude from AC to B is up in the original and left in the first image, indicating a rotation 90° CCW.

The first transformation is a rotation 90° CCW.

The vertex order of A''B''C'' is CCW, indicating a reflection. The direction of the altitude from A''C'' to B'' is still to the left, so the reflection must be over a horizontal line. We find the x-axis bisects the segments A'A'', B'B'', and C'C'', confirming that it is the line of reflection.

The second transformation is reflection across the x-axis.

_____

Algebraically, the transformations are ...

1st: (x, y) ⇒ (-y, x)

2nd: (x, y) ⇒ (x, -y)

Both together: (x, y) ⇒ (-y, -x).

6 0

3 years ago

60°<br> 4V3<br> Find the missing side

Vinvika [58]

Check the picture below.

7 0

4 years ago

Prove that 1³+2³+....n³=n²(n+1)²/4. principle of mathematics induction

alex41 [277]

<h3>The Simplified Question:-</h3>

$\sf 1^3+2^3\dots n^3=\dfrac{n^2(n+1)^2}{2}$

$\\ \sf\longmapsto 1^3+2^3+\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

<h3>Solution:-</h3>

Let

$\\ \sf\longmapsto P(n)=1^3+2^3\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

For n=1

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(1+1)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(2)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{2}{2}\right)^2$

$\\ \sf\longmapsto P(1)=(1)^2$

$\\ \bf\longmapsto P(1)=1=1^3$

Let k be any positive integer.

$\\ \sf\longmapsto P(k)= 1^3+2^3\dots k^3=\left(\dfrac{k(k+1)}{2}\right)^2$

We have to prove that p(k+1) is true.

consider

$\sf 1^3+2^3\dots k^3+(k+1)^3$

$\\ \sf\longmapsto \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2}{4}+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2+4(k+1)^3}{4}$

$\\ \sf\longmapsto \dfrac{k+1)^2\left\{k^2+4k+4\right\}}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+2)^2}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+1+1)^2}{4}$

$\\ \sf\longmapsto \left(\dfrac{(k+1)(k+1+1)}{2}\right)^2$

$\\ \sf\longmapsto (1^3+2^3+3^3\dots k^3)+(k+1)^3$

Thus P(k+1) is true whenever P(k) is true.

Hence by the Principal of mathematical induction statement P(n) is true for $\bf n\epsilon N.$

Note:-

We can solve without simplifying the Question .I did it for clear steps and understanding .

<h3>Learn More:-</h3>

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It is y- that means a

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68 students increased by 125%

kramer

68 students increased by 125% is

68 + 68(1.25)

68 + 85 = 153

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