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loris [4]
4 years ago
11

That Is all l need to know

Mathematics
2 answers:
nikitadnepr [17]4 years ago
8 0
Pick a number and write it 4 times to make it a 4 digit number. (example: 7,777.
7 is the number I chose)
No matter what number you chose the number farthest to the left would be the largest number.(the number you would circle)
Elza [17]4 years ago
6 0
5,555 and the 5 and the beging would be circled because of the amount of other numbers there are the first five is not just five it is 5,000
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Cual es la mediana de los datos en un estudio​
seraphim [82]

Answer:

La mediana de un conjunto de números es el número del medio del conjunto (después de que los números se hayan ordenado de menor a mayor).

Step-by-step explanation:

Digamos que tienes estos números, 4, 7, 2, 9, 7, 6 y 4. primero, los pones en orden de menor a mayor (2, 4, 4, 6, 7, 7, 9), y luego usted determina qué número está en el medio, que es 6.

3 0
3 years ago
translate the sentence into a equation using n as the unknown number . Then solve the equation for n . 5 increased by half a num
Ksju [112]

Answer:

I think it'll be 5 + n = 11

11 - 5 = 6

n = 6

5 0
3 years ago
Read 2 more answers
Given the equation y=x^2-12x-5, convert to vertex form, then give the value of k
marishachu [46]

the answer os 5xx entendida

6 0
3 years ago
the circumference of a circle is c centimeters the diameter of the circle is 13 centimeters what expression best represents the
Blababa [14]

Answer:

C/13 = pi

Step-by-step explanation:

7 0
3 years ago
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
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