All categories

3 years ago

PLEASE HELP ASAP!!!!!

Mathematics

1 answer:

dezoksy [38]3 years ago

4 0

I think 4 most likely because if u take the middle of the 3 dots it would lave the last dot on the 4

You might be interested in

Plz help, And no links please

Talja [164]

I did not get any zeros since the graph doesn’t cross the x axis, meaning that there are no rational zeros

However, here is the method u can use to find the zeros lol

You can use the quadratic formula in order to get the zeros

This is the equation therefore use these values
ax^2+bx+c=0

A=1
B= -5
C=12

The quadratic formula is -b±√(b^2-4ac))/2a (I left a picture just in case)

6 0

3 years ago

Evaluate x³ for x=2.

loris [4]

Answer:

Step-by-step explanation:

If we have anything to the third power, we are multiplying the number by itself 3 times.

If x = 2, then the expression is $2^3$ .

$2\cdot2\cdot2=8$

Hope this helped!

7 0

3 years ago

Read 2 more answers

What is the largest divisor of 342 that is also a factor of 285?

Scilla [17]

Answer:57

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir

kati45 [8]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
\begin{cases}
\stackrel{center}{(-1,0)}\\
\stackrel{radius}{6}
\end{cases}$

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}
\\\\\\
d=\sqrt{0+1}\implies d=1$

well, the distance from the center to A is 1, namely is "inside the circle".

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2}
\\\\\\
d=\sqrt{0+36}\implies d=6$

notice, the distance to B is exactly 6, and you know what that means.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})
\\\\\\
\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}
\\\\\\
d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398$

notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.

3 0

3 years ago

Read 2 more answers

Someone help me please

lina2011 [118]

Answer:

-13, 0.4,7 its the 4th one

Step-by-step explanation:

3 0

3 years ago