To add 11% to an amount means to multiply the amount by 1.11.
You can start with $115 and multiply it by 1.11 eleven times.
A quicker way is to find 1.11^11 and then multiply by $115.
$115 * 1.11^11 = $115 * = 3.151757 = $362.45
The quotient of z and 3.
quotient = ÷
z/3
The answer is A.
Check the picture below.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -1}}\quad ,&{{ 2}})\quad % (c,d) &({{ 2}}\quad ,&{{ 4}})\\ &({{ 2}}\quad ,&{{ 4}})\quad % (c,d) &({{ 3}}\quad ,&{{ -2}})\\ &({{ 3}}\quad ,&{{ -2}})\quad % (c,d) &({{ -2}}\quad ,&{{ -3}})\\ &({{ -2}}\quad ,&{{ -3}})\quad % (c,d) &({{ -1}}\quad ,&{{ 2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-1%7D%7D%5Cquad%20%2C%26%7B%7B%202%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%202%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5C%5C%0A%26%28%7B%7B%202%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%203%7D%7D%5Cquad%20%2C%26%7B%7B%20-2%7D%7D%29%5C%5C%0A%26%28%7B%7B%203%7D%7D%5Cquad%20%2C%26%7B%7B%20-2%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%5C%5C%0A%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-1%7D%7D%5Cquad%20%2C%26%7B%7B%202%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D)
![\bf -------------------------------\\\\ d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2} \\\\\\ d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\ -------------------------------\\\\ d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\ -------------------------------\\\\ d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2} \\\\\\ d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%284-2%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%282%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B3%5E2%2B2%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B13%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%283-2%29%5E2%2B%28-2-4%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B1%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B37%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-2-3%29%5E2%2B%5B-3-%28-2%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-5%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
![\\\\ -------------------------------\\\\ d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2} \\\\\\ d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-1-%28-2%29%5D%5E2%2B%5B2-%28-3%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B2%29%5E2%2B%282%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%281%29%5E2%2B%285%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
so, those are their lengths, sum them all up, that's the polygon's perimeter.
Answer:
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Step-by-step explanation:
Answer:
the answer is 0.1 dime
Step-by-step explanation:
2 1/2 % = 2.5%
2.5% of 4 = 4 * 2.5/100
= 0.1 dime
Amount left with Mathew = 4 - 0.1 = 3.9 dimes
Mathew has 3.9 dimes left with him.