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3 years ago

What is 2848448 X 39494948

Mathematics

2 answers:

Greeley [361]3 years ago

6 0

Answer:

1.12499306E14

Step-by-step explanation:

i did the math

maw [93]3 years ago

3 0

Answer:

112,449,305,640,704

Step-by-step explanation:

2,848,448 * 39,494,948 = x

2,848,448 * 39,494,948 = 112,449,305,640,704

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The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 15.4 V and b = -5.10 V/m.

Minchanka [31]

Answer:

$V(0)=15.4\\V(3)=0.1\\V(6)=-15.2$

Step-by-step explanation:

The potential in a region between x = 0 and x = 6.00 m is V = a + bx.

Put a = 15.4 V and b = -5.10 V/m.

So, $V=15.4-5.10x$

To find: Potential at x =0, 3, 6

Solution:

Potential at x = 0:

$V=15.4-0=15.4$

Potential at x = 3:

$V=15.4-5.10(3)=15.4-15.3=0.1$

Potential at x = 6:

$V=15.4-5.10(6)=15.4-30.60=-15.2$

Therefore,

$V(0)=15.4\\V(3)=0.1\\V(6)=-15.2$

7 0

4 years ago

The midpoint of GĦ is M(-3, 4). One endpoint is G(-2, 2). Find the<br> coordinates of endpoint H.

prisoha [69]

Answer:

Step-by-step explanation:

(x-2)/2 = -3

x - 2 = -6

x = -4

(y + 2)/2 =4

y + 2 =8

y = 6

(-4, 6)

7 0

4 years ago

En los siguientes ejercicios, encuentra la probabilidad 1) Que no salga ningun 4 al tirar dos dados.

LenaWriter [7]

Answer:

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7 0

3 years ago

Write standard form of line that is parallel to 2x + 3y= 4 and passes through the point (1, -4)

Tresset [83]

Answer:

2x + 3y = -10

Step-by-step explanation:

First convert the line into slope intercept form to find the slope of the line.

2x + 3y = 4

3y = 4 - 2x

y = -2/3x + 4/3

The slope of the line is -2/3. Parallel lines have the same slope so the slope for a line through the point (1,-4) will be -2/3. Substitute m = -2/3 and (1,-4) into the point slope of a line.

$y --4 = -\frac{2}{3}(x-1)\\y + 4 = -\frac{2}{3}(x -1)$

Now convert the line into standard form by using the distributive property.

$y + 4 = -\frac{2}{3}(x -1)\\y + 4 = -\frac{2}{3}x + \frac{2}{3}\\3y + 12 = -2x + 2\\2x + 3y + 12 = 2\\2x + 3y = -10$

4 0

4 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago

What is 2848448 X 39494948​

What is 2848448 X 39494948