HELP Ill give brainlist

Which best describes her prediction?

ohaa [14]

There is no picture or anything but it’s ok .

3 0

3 years ago

PLEASE HELP ME WITH THIS. 20 POINTS

Gelneren [198K]

Answer:

y=12.75x and y=13x

Step-by-step explanation:

The rate of change is another name for slope. To find the slope of Relationship B, we use the formula

m = \frac{y_2-y_1}{x_2-x_1}

Using the first two points, we have

m = (50-25)/(4-2) = 25/2 = 12.5

This means that anything with a rate of change greater than 12.5 works in this problem.

The equations listed for Relationship A are in slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept (in all of these, b = 0).

In the first equation, m = 13; in the second equation, m = 11.5; in the third equation, m = 12.75; and in the fourth equation, m = 12.25. The only ones with higher slopes than Relationship B are the first and the third one.

4 0

3 years ago

Read 2 more answers

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where h and k forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .