F=\dfrac{9}{5}C+32\\\\A.\\\dfrac{9}{5}C+32=F\ \ \ |-32\\\\\dfrac{9}{5}C=F-32\ \ \ |\cdot\dfrac{5}{9}\\\\C=\dfrac{5}{9}(F-32)
B.\\F=212\to C=\dfrac{5}{9}(212-32)=\dfrac{5}{9}\cdot180=100\\\\212^oF=100^oC
C.\\F=80\to C=\dfrac{5}{9}(80-32)=\dfrac{5}{9}\cdot48\approx26.7\\\\80^oF\approx26.7^oC
Answer:
0.10
Step-by-step explanation:
Technically it's 0.12 but I only say this because they rounded by from 58 to 60 in their question
Answer:
In first equation if we put the value of x and y (x=2,y=2)than,
3x+y=8
3×x+y=8
3×2+2=8
6+2=8
8=8
in second equation the value of x=2,y=1
x2+xy=6
x2+x×y=6
4+2×1=6
4+2=6
6=6
The width used for the car spaces are taken as a multiples of the width of
the compact car spaces.
Correct response:
- The store owners are incorrect
<h3 /><h3>Methods used to obtain the above response</h3>
Let <em>x</em><em> </em>represent the width of the cars parked compact, and let a·x represent the width of cars parked in full size spaces.
We have;
Initial space occupied = 10·x + 12·(a·x) = x·(10 + 12·a)
New space design = 16·x + 9×(a·x) = x·(16 + 9·a)
When the dimensions of the initial and new arrangement are equal, we have;
10 + 12·a = 16 + 9·a
12·a - 9·a = 16 - 10 = 6
3·a = 6
a = 6 ÷ 3 = 2
a = 2
Whereby the factor <em>a</em> < 2, such that the width of the full size space is less than twice the width of the compact spaces, by testing, we have;
10 + 12·a < 16 + 9·a
Which gives;
x·(10 + 12·a) < x·(16 + 9·a)
Therefore;
The initial total car park space is less than the space required for 16
compact spaces and 9 full size spaces, therefore; the store owners are
incorrect.
Learn more about writing expressions here:
brainly.com/question/551090
The slope is 3/1
hope this helps
