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3 years ago

11^-5 ÷ 11^-3 please help!!!!

Mathematics

2 answers:

levacccp [35]3 years ago

7 0

Answer:

11^ -2 = 1 / 11^2 = 1/121

Step-by-step explanation:

11^-5 ÷ 11^-3

When the bases are the same, we subtract the exponents when dividing

11 ^ ( -5 - -3)

11 ^ ( -5+3)

11 ^ -2

We know that a^ -b = 1/a^b

1/ 11^2

1/121

prohojiy [21]3 years ago

6 0

Answer:

Step-by-step explanation:

$a^{-m}=\frac{1}{a^{m}}\\\\Therefore, \frac{1}{a^{-m}}=a^{m}$

$11^{-5}$ ÷ $11^{-3} = 11^{-5} * 11^{3}\\\\$ { $a^{m}*a^{n} = a^{m+n}$ }

$11^{-5+3} \\\\= 11^{-2}$

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Which situation CANNOT be represented by this equation? 6x−7=29<br>

kotykmax [81]

Answer:

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Step-by-step explanation:

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3 years ago

(?,5) is on the line 4x-7y=1. Find the other half of the coordinate.

ludmilkaskok [199]

The coordinate is an ordered pair $(x,y)$ . We have a number, 5, for y, which we can substitute (plug-in) for y in the equation of the line and solve for x.

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3 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:

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

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This can be rewritten as

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

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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

1. Ana was washing dishes at the end of her shift at a restaurant. She washed 30 plates from 7:30 to 7:35. When she finishes was

irga5000 [103]

A)

since she washed 30 plates in 5 mins (7:30 to 7:35) the ratio was 5:30 or in simplest form 1min:6plates

This means she washes 6 plates a min and thus from 7:15 to 7:38 (23 mins) she washes:

6X23 = 138 plates

b)

L - plates left
T- total plates
t - time(mins)

"L = T - 6x"
This is because she washes 6 plates a min

7 0

3 years ago

A potato was launched in the air using a potato gun. The function for the situation was h(t)=-16t^2 +100t+10, where t was the ti

Goshia [24]

Answer:

The height equation is:

h(t)=-16t^2 +100t+10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = -2*16*t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest heigth of the potato will be h = 0ft, when the potato hits the ground.

C) the lowest time that works for that function is t = 0s, when the potato is fired by the gun.

D) the maximum time can be finded when the potato hits the ground, afther that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

$t = \frac{-100 +- \sqrt{100^2 +4*16*10} }{-2*16} = \frac{-100 +- 103.2}{-32}$

So the two solutions are:

t = 6.3s

t = -0.1s

We need to chose the positive time, as we already discused that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is 0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

4 0

3 years ago

11^-5 ÷ 11^-3 please help!!!!​

11^-5 ÷ 11^-3 please help!!!!