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3 years ago

Which statements about the graph of the function y=(1/3)

1 answer:

Elena-2011 [213]3 years ago

8 0

Answer: The range of the function is all real numbers is true.

Explanation: the graph of y = 1/3 is just a horizontal line at y = 1/3. so it’s not increasing or decreasing (it’s staying the same). there is no x-intercept, and the y-intercept is (0, 1/3). the range is all the possible y values, and there are infinite possible y values, so it is all real numbers.

hope this helps!

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Answer: (1,3)

Explanation:

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3 years ago

(3square root8x^5)-(4square root256y^3) Write in exponential form and simplify

lana66690 [7]

Answer:

$2x^\frac{5}{3} - 4y^\frac{3}{4}$

Step-by-step explanation:

Given equation:

$\sqrt[3]{8x^5} - \sqrt[4]{256y^3}$

( $(8x^5)^\frac{1}{3} - (256y^3)^\frac{1}{4} \\ (8^\frac{1}{3} . x^\frac{5}{3} ) - (256^\frac{1}{4}. y^\frac{3}{4})\\ 2x^\frac{5}{3} - 4y^\frac{3}{4}$ !

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4 years ago

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4 years ago

Question 9

Alenkasestr [34]

Answer:

$PX=4\\PY=4\sqrt{3}$

Step-by-step explanation:

Let $\angle PYZ=Q$

$\Rightarrow \angle XYP=90-\angle PYZ\\\angle XYP=90-\theta$

In the $\Delta PYZ$

$\tan\theta =\frac{opposite}{adjacent}=\frac{PZ}{PY}......(1)$

In the $\Delta YPX$

$\tan(90-\theta)=\frac{opposite}{adjacent}=\frac{PX}{PY}$

$\cot=\frac{PX}{PY}....(2)$

eqn(1) $\times$ eqn(2)

$\tan\times\cot\theta=\frac{PZ}{PY}\times\frac{PX}{PY}\\\\1=\frac{PZ\times PX}{PY^2}\ \ (as\ tan\theta\ =\frac{1}{\cot\theta})\\\\PY^2=PZ\times PX\\PY^2=12PX.......(3)$

Now in $\Delta XYP$

use Pythagorean theorem

$XY^2=PY^2+PX^2\\8^2=PY^2+PX^2\ \ (as\ XY\ =8)\\PX^2+PY^2=64\\\\PY^2=12PX\ \ (eqn(3)\\\\\Rightarrow PX^2+12PX=64\\\\PX^2+12PX-64=0\\\\PX^2+16-4PX-64=0\\\\(PX+16)(PX-4)=0\\\\PX=4,\ -16$