You have to find the area of both the square and rectangle. use the formula A=wl
square = 3x3 = 9
rectangle = 10x3 = 30
area= 39 (i think)
Answer:
Step-by-step explanation:
Let's do the obvious things first.
Right rectangle
w = 11
L = 17
Formula
Area = L * W
Area = 11 * 17
Area = 187
Left rectangle
W = 8
L = 11
Area = 8 * 11= 88
Middle Rectangle
W = 11
L = 15
Area = 11 * 15 165
Now we come to the triangles. It's not obvious what to do with them. You have to infer that one of the lets is 8 and the other is 15. You get the 8 from the width of the left rectangle (see above.)
There are 2 triangles
Area = 2 * 1/2 * l1 * l2
Leg 1 = 8
Leg2 = 15
Area = 2 * 1/2 (8 * 15)
Area = <u> 120 </u>
Total = 120 + 165 + 88 + 187 = 560
Unless you have an E answer that is 560, my answer does not agree with any of the given answers. I would suggest that you ask your teacher how it is done. My method is correct.
Answer: 2x+3y=18
Step-by-step explanation: Any line parallel to 2x+3y+11=0 is the form 2x+3y =k this can also be written as x/(k/2)+y/k/3 =1 this is the line whose sum of the intercepts is 15, then k/2 +k/3 = 15 after solving we get that indeedk= 18.
First,
149,500 * .10 = 14,950
149,500 - 14,950 = 134,550
then,
134,550 * .05 = 6727.50
134,550 - 6727 = 127,822.55
127,822.50 * .04 = 5112.90
trade discount = 14950 + 6727.50 + 5112.90
26790.40
hope this help
Answer:
Option (d) is correct.
Step-by-step explanation:
Given : Expression 
We have to write a simplified form of the given expression
Consider the given expression
![\mathrm{Apply\:radical\:rule\:}\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%5C%3A%7D%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
Factor 10000 as 
![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da)
also, ![\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%5C%3A%7D%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
We have,
Thus,
