The answer is 5
a^2+b^2=c^2
9+16=25
sq root of 25 is 5
Answer:
5.9125in^2
Step-by-step explanation:
Step one:
Given data
Dimension of square= 5.25 in
Area of sqaure= 5.25^2= 27.5625 in^2
We are told that the sides of the circle touch the sides of the square
hence the diameter of the circle is 5.25in
radius= 5.25/2= 2.625 in
Area of circle= πr^2
Area of circle = 3.142*2.625^2
Area of circle=3.142*6.890625
Area of circle= 21.65 in^2
Step two:
The area that is not inside the circle
=Area of square-Area of circle
=27.5625- 21.65
=5.9125in^2
What you want to do here is take this information and plug it into point-slope form. any time you're given a point and a slope, you generally want to plug it into this equation: y - y1 = m(x - x1).
in this equation, m is your slope and (x1, y1) is a given point. plug in your info--slope of -3 and (-5, 2).
y - 2 = -3(x + 5)
that is the equation of your line. however, if you want to graph it, this doesn't really make much sense to you. convert it to slope-intercept form, y = mx + b, by solving for y.
y - 2 = -3(x + 5) ... distribute -3
y - 2 = -3x - 15 ... add 2
y = -3x - 13 is your equation.
to graph this, and any other y = mx + b equation, you want to start with your y-intercept if it's present. your y intercept here is -13, which means the line you wasn't to graph crosses the y-axis at y = -13, or (0, -13). put a point there.
after you've plotted that point, you use your slope to graph more. remember that your slope is "rise over run"--you rise up/go down however many units, you run left/right however many units. if your slope is -3, you want to go down 3 units, then go to the right 1 unit. remember that whole numbers have a 1 beneath them as a fraction. -3/1 is your rise over 1.
SOLUTION
Given B is the midpoint of AC ABBC 1 and C is the midpoint of BD BCCD2 ie on adding 1 and 2 ART We get ABBCBCCD hence ABCD hence proved
Answer: (4.9x) - 3
Step-by-step explanation: the reason i didn't add an extra line thing is because the negative becomes the minus. the 4.9 is self evident
