All categories

3 years ago

Please answer these 2 questions

Mathematics

2 answers:

charle [14.2K]3 years ago

4 0

1. Perpendicular to its radius
2. 0

iren2701 [21]3 years ago

4 0

Yes the first person is right.

You might be interested in

A line passes through the points (2,18) and (8,24). Write a linear function rule in terms of x and y for this line. The linear f

Alinara [238K]

Answer:

Step-by-step explanation:

hello : here is an solution

8 0

3 years ago

Point A is located at (2, 6), and D is located at (−4, 10). Find the coordinates of the point that lies halfway between A and D.

valina [46]

$(( \frac{1}{2} (( - 4) - 2) + 2and( \frac{1}{2} (10 - 6)) + 6$
the answer is(-1,8)

8 0

3 years ago

Read 2 more answers

How many triangles can be constructed with 3 angles, each measuring 63°?

earnstyle [38]

Answer:

a triangle's angles should add up to be 180 degrees. Three angles of 63's sum is 189

4 0

3 years ago

Read 2 more answers

How to determine if an outcome is unusual

lbvjy [14]

The only way to determine whether or not an outcome is unusual or not is to compare it with previous or future outcomes.

3 0

3 years ago

Find a and b if the point p(6,0) and Q(3,2) lie on the graph of ax+ by=12

tiny-mole [99]

to get the equation of any straight line we only need two points off of it, hmmm let's use P and Q here and then let's set the equation in standard form, that is

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$(\stackrel{x_1}{6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{6}}}\implies \cfrac{2}{-3}\implies -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{6})$

$\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-0)~~ = ~~3\left( -\cfrac{2}{3}(x-6) \right)}\implies 3y=-2(x-6) \\\\\\ 3y=-2x+12 \implies \stackrel{a}{2} x+\stackrel{b}{3} y=12$

5 0

2 years ago