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3 years ago

If you think of the equation as an equation for a line.

Mathematics

1 answer:

anzhelika [568]3 years ago

7 0

Answer:

The final answer is D

Step-by-step explanation:

Easy.

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Who won the race between the two balls of string

motikmotik

The higher one I think

7 0

3 years ago

Which is the graph of f (x) = 4 (1/2) Superscript x?

miss Akunina [59]

Answer:

Option (2) is correct graph.

Step-by-step explanation:

Given:

The function to graph is given as:

$f(x)=4(\frac{1}{2})^x$

Now, the above function is an exponential function of the form $f(x)=ka^x$

Where, 'k' and 'a' are constants.

The range of an exponential function is always greater than 0.

The domain is all real numbers.

Now, for graphing it, we need to find some points on it and its end behaviour.

Now, for x = 0, the function value is given as:

$f(0)=4(\frac{1}{2})^0=4$

So, (0, 4) is a point on the graph.

Now, for x = 1, the function value is given as:

$f(1)=4(\frac{1}{2})^1=4\times\frac{1}{2}=2$

So, (1, 2) is another point on the graph.

Now, for x = 2, the function value is given as:

$f(2)=4(\frac{1}{2})^2=4\times\frac{1}{4}=1$

So, (2, 1) is another point on the graph.

Now, as 'x' tends to ∞, the function value tends to:

$f(x\to\infty)=4\cdot\frac{1}{2}^{\infty}=\frac{4}{\infty}=0$

So, as

$x\to\infty,f(x)\to0\\\\x\to-\infty,f(x)\to\infty$

Now, from among all the options, only option (2) fulfills all the conditions given above.

So, option (2) is correct graph.

5 0

4 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$