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3 years ago

Use the graph of the function y = 4x to answer the following questions.

Mathematics

2 answers:

AURORKA [14]3 years ago

5 0

Answer:

The domain of the function is negative infinity to positive infinity because an exponent can be any real number.

Sav [38]3 years ago

4 0

Hello from MrBillDoesMath!

Answer:

For y = 4x, the domain of the function is all real numbers. This can be indicated as ( -infinity, infinity),

...later part of question:...

..."because an exponent can be any real"

exponent? !!!

Did you mean y = 4^x ? In this case the domain of y is all real numbers ann the range is (0, infinity), where the value 0 is not included in the range.

Thank you,

MrB

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On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$