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3 years ago

6

Simplity (4^-2)^5 A.1/4^10 B. 4^3 C.1/4^3 D. 4^10

2 answers:

Goshia [24]3 years ago

5 0

Answer:

$(4^{-2})^5=\dfrac{1}{4^{10}}$

Step-by-step explanation:

The given expression is : $(4^{-2})^5$

We need to simplify the above expression.

We know that, $x^{-a}=\dfrac{1}{x^a}$

or

$(4^{-2})^5=(\dfrac{1}{4^2})^5\\\\=\dfrac{1^5}{(4^2)^5}\\\\\because (x^b)^c=x^{b\times c}\\\\=\dfrac{1}{4^{10}}$

So, the simplified form of the given expression is $\dfrac{1}{4^{10}}$ . Hence, the correct option is (A).

andriy [413]3 years ago

3 0

Answer:

The answer is a 1/4^ 10.

Step-by-step explanation:

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3 years ago

Two collinear points on a line are given in the table below:

katrin [286]

Answer:

$(4,3)$ and $(7,2)$ do not lie on the line

Step-by-step explanation:

Given

$(0,0)\ and\ (2,1)$

Required

Determine which points that are not on the line

First, we need to determine the slope (m) of the line:

$m = \frac{y_2 - y_1}{x_2- x_1}$

Where

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$(x_2,y_2) = (2,1)$

So;

$m = \frac{y_2 - y_1}{x_2- x_1}$

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Next, we determine the line equation using:

$y - y_1 = m(x -x_1)$

Where

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$(x_1,y_1) = (0,0)$

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$y - 0 = \frac{1}{2}(x - 0)$

$y = \frac{1}{2}x$

To determine which point is on the line, we simply plug in the values of x to in the equation check.

For $(4,2)$

$x = 4$ and $y =2$

Substitute 4 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 4$

$2 = \frac{4}{2}$

$2=2$

<em>This point is on the graph</em>

<em></em>

For $(4,3)$

$x = 4$ and $y = 3$

Substitute 4 for x and 3 for y in $y = \frac{1}{2}x$

$3 = \frac{1}{2} * 4$

$3 = \frac{4}{2}$

$3 \neq 2$

<em>This point is not on the graph</em>

<em></em>

For $(7,2)$

$x = 7$ and $y = 2$

Substitute 7 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 7$

$2 = \frac{7}{2}$

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<em></em>

<em>This point is not on the graph</em>

<em></em>

<em></em> $(\frac{4}{8},\frac{2}{8})$ <em></em>

<em></em> $x = \frac{4}{8}$ and<em> </em> $y = \frac{2}{8}$ <em></em>

<em>Substitute </em> $\frac{4}{8}$ <em> for x and </em> $\frac{2}{8}$ <em> for y in </em> $y = \frac{1}{2}x$ <em></em>

<em></em> $\frac{2}{8} = \frac{1}{2} * \frac{4}{8}$ <em></em>

<em></em> $\frac{2}{8} = \frac{1 * 4}{8 * 2}$ <em></em>

<em></em> $\frac{2}{8} = \frac{4}{16}$ <em></em>

<em></em> $\frac{1}{4} = \frac{1}{4}$ <em></em>

<em></em>

<em>This point is on the graph</em>

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Answer:

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The volume's rate of change is written as a function of time.

- $\frac{dh}{dt}$ is the rate of change in the height of water in the tank with respect to time, t.

b. $\frac{dV}{dt}-$ is the only constant. Water flows into the constant at a constant rate, say $6cm^3$ per minute.

c. $\frac{dV}{dt}$ is positive. Volume water in the take is increasing from time to time.

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Answer:

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Alex17521 [72]

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Read 2 more answers