Answer: a
Step-by-step explanation:
It’s B if there are straight lines and B is there dot it but did u post the same question two times?
The actual width of the room is 20 ft and the actual length of the room is 15 ft
Since on the scale, 2 in : 5 ft and the width of the room on the drawing is 8 in.
Let the actual width of the room is w.
The ratio of the drawing to actual width is 8 in : w
So, 2 in : 5 ft = 8 in : w
2 in/5 ft = 8 in/w
So, w = 8 in × 5 ft/2 in
w = 4 × 5 ft
w = 20 ft
Also, the length of the room on the drawing on the drawing is 6 in.
Let the actual length of the room is L.
The ratio of the drawing to actual length is 6 in : L
So, 2 in : 5 ft = 6 in : L
2 in/5 ft = 6 in/L
So, L = 6 in × 5 ft/2 in
L = 3 × 5 ft
L = 15 ft
So, the actual width of the room is 20 ft and the actual length of the room is 15 ft.
Learn more about scale drawing here:
brainly.com/question/25324744
Subtract 1111 from both sides
5{e}^{{4}^{x}}=22-115e4x=22−11
Simplify 22-1122−11 to 1111
5{e}^{{4}^{x}}=115e4x=11
Divide both sides by 55
{e}^{{4}^{x}}=\frac{11}{5}e4x=511
Use Definition of Natural Logarithm: {e}^{y}=xey=x if and only if \ln{x}=ylnx=y
{4}^{x}=\ln{\frac{11}{5}}4x=ln511
: {b}^{a}=xba=x if and only if log_b(x)=alogb(x)=a
x=\log_{4}{\ln{\frac{11}{5}}}x=log4ln511
Use Change of Base Rule: \log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}logbx=logablogax
x=\frac{\log{\ln{\frac{11}{5}}}}{\log{4}}x=log4logln511
Use Power Rule: \log_{b}{{x}^{c}}=c\log_{b}{x}logbxc=clogbx
\log{4}log4 -> \log{{2}^{2}}log22 -> 2\log{2}2log2
x=\frac{\log{\ln{\frac{11}{5}}}}{2\log{2}}x=2log2
Answer= −0.171
Answer:
8
Step-by-step explanation:
If you add x to the left side of the equation you get positive 1/2x +3=7
you then would subtract 3 from 7 to get 4
this would leave you with 1/2x=4
if you divide 4 by 1/2 you get 8 as the answer.
