All categories

2 years ago

What are the zeros of this function?

Mathematics

2 answers:

miss Akunina [59]2 years ago

5 0

Answer:

A. x = 3 and x = 6

Step-by-step explanation:

Zeros occur when the function crosses the x - axis. In this case, the quadratic function crosses the function when x = 3 and x = 6.

tatyana61 [14]2 years ago

3 0

Answer:

Step-by-step explanation:

the zeroes of a function basically mean when y = 0, so basically the x-intercept(s)

in this case, the zeroes are 3 and 6

You might be interested in

Write a trinomial which is degree 2, where the coefficient of the highest degree term is 7.

Nady [450]

Answer:

7x^2 - 2x - 3.

Step-by-step explanation:

Starts with 7x^2 then a term in x then a constant.

4 0

2 years ago

0.4 x 10 =....<br>1.25 x 10 = <br>97.465 x10 =

chubhunter [2.5K]

Answer:

0.4 × 10 = 4

1.25 × 10 = 12.5

97.465 × 10 = 974.65

3 0

3 years ago

A pen cost y dollars,how much will 25 such pens cost

kiruha [24]

Answer:

25y

Step-by-step explanation:

y×25=25y

3 0

2 years ago

What the constant rate of change?

lozanna [386]

For straight lines the constant rate of change(slope) is a constant(always the same) For every unit moved on the x-axis two more are moved on the y-axis

6 0

2 years ago

Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)

marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}$

8 0

2 years ago