All categories

2 years ago

If a graph passes the horizontal line test, but not the vertical line test, is it still a function?

Mathematics

1 answer:

nydimaria [60]2 years ago

8 0

If a graph fails the vertical line test it's not a function; you can't have a single x map to two ys.

Vice versa, if the graph passes the verticle line test, it's a function.

The horizontal line test tests for injectivity aka one-to-one-ness; it has no bearing on whether something's a function.

You might be interested in

Please help with this square problem, the picture is shown.

Galina-37 [17]

<span>Area of the upper rectangle = x(16 - 6) = 10x </span>yd²<span>

Area of the bottom rectangle = 6(x + 6) = 6x + 36 </span>yd²<span>

Total area = 10x + 6x + 36 = 16x + 36 yd</span>²

5 0

3 years ago

|x|-2≤3<br>|x+1| +5<7 how do I shove this

Fofino [41]

$3. < \leqslant {?}^{2}$

6 0

3 years ago

Find the volume of the cylinder if the height is 3 mm and the diameter is 40 mm. Leave the answer in terms of it.

KiRa [710]

The Volume of cylinder is 3770 cubic millimeters

But, the correct option is not available in options given.

Step-by-step explanation:

Height of cylinder = 3mm

Diameter of cylinder = 40mm

Volume of cylinder = ?

The formula used to find Volume of cylinder is:

$Volume = \pi r^2h$

Putting values in the formula:

Since we need radius,

radius = diameter/2

radius = 40/2

radius = 20 mm

So,

$Volume = \pi r^2h\\Volume = 3.14*(20)^2*3\\Volume = 3.14*400*3\\Volume= 3769.91 or \,\,\,Volume\approx 3770\,\,cubic\,\,millimeters$

The Volume of cylinder is 3770 cubic millimeters

But, the correct option is not available in options given.

Keywords: Volume of cylinder

Learn more about Volume of cylinder at:

#learnwithBrainly

6 0

3 years ago

Marissa was decorating her room .she arranged 63 same size picture tiles on a wall in the shape of a rectangle.which are possibl

xxMikexx [17]

21 by 3 because 63÷3=21

8 0

2 years ago

Read 2 more answers

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

2 years ago