I need info on Albert Einstein

2 answers:

4 0

<span>Born on March 14, 1879 in Ulm, Württemberg, Germany, Albert Einstein grew up in a secular Jewish family. ... Einstein had one sister, Maja, born two years after him.Einstein attended elementary school at the Luitpold Gymnasium in Munich.</span>

Slav-nsk [51]3 years ago

3 0

Just use wiki! I tried to post the website, but it removes it! Sorry!

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Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter

Alex Ar [27]

Answer:

I would need more information to help you

Step-by-step explanation:

7 0

3 years ago

Help help math math math

Marianna [84]

Answer:

Step-by-step explanation:

Get rid of fractions:

Multiply the whole equation, i.e. all terms both sides of the equals sign, by the denominator or the lowest common multiple of all denominators if there are multiple fractions in the equation;

In this case, only one term is a fraction and therefore has a denominator (i.e. 7), so it is this number we multiply the equation by to get:

7(y) = 7(-³/₇.x) + 7(3)

7y = -3x + 21

3x + 7y = 21

5 0

2 years ago

Example 2:

Montano1993 [528]

Taking this example into account, we can see that setting the first value equal to 1, we obtain that $F(x)=0.5x+1=4$ and $x=6$ . Using this information, we find that $F(x+1)=0.5(x+1)+1=0.5(6+1)+1=4.5$ . It shows that when x is positive, the succussive terms are increasing.

Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that $F(x)=0.5x+1=-19$ and x=-40. In the next iteration, $F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5$ . In the next iteration, $F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18$ . By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.

Using the function $g(x)=-x+2$ and taking the initial value equal to 4, we find that $g(x)=-x+2=4$ and x=-2. In the next iteration, $g(x+1)=-(x+1)+2=-(-2+1)+2=3$ . If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that $g(x)=-x+2=2$ and x=0. The next iteration is $g(x+1)=-(x+1)+2=1$ . In this case, the interations are also decreasing.
If we set the initial value equal to 1, we find that $g(x)=-x+2=1$ and x=1. In the next iteration, $g(x+1)=-(x+1)+2=0$ and the iterations are decreasing.