The difference = miles rode on Tuesday - miles rode on Thursday.
6 5/6 - 5 5/6 = 1 mile
The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11 
⇒x=11t
Area of square=
Area of square=
{as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
Learn more about The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11
⇒x=11t
Area of square=
Area of square={as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
Learn more about area here:
brainly.com/question/27683633
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For the first one, you have to convert the fractions to an improper fraction. To do that you need to multiply the bottom denominator number (3) by the whole number (1) then you need to add the numinator, so 3x1+2= 5. You have to keep the denominator so 1 2/3 is equal to 5/3. Then do the same to 2 1/5, and you get 11/5. Now you have to find a common denominator, that's basically the smallest number that both numbers can go Into, the lowest common denominator for 3 and 5 is 15. So 3x5= 15, so we have to multiply the top number by 5 which is 25. So 5/3 is equal to 25/15, then 5x3= 15, so you need to multiply 11 by 3 which is 33. So 11/5 is equal to 33/15. Then you add them. Add the numinators (25+33=58. Then you keep the denominator 15. So when u add it it's 58/15 then you need to simplify that and you get 3 13/15.
The second one you turn them into improper fractions like I told you how to before (multiply the bottom number by the whole number then add he top number, then add he same denominator.) do that for both. Then you line them right next to each other and multiply across. (I just realized that they were the same number so they are equal to 5/3 and 11/5)
Then you do 5x11 and you get 55 then do 3x5 and you get 15. 55/15 is your answer, but you need to simplify it, you need to divide 55 by 15, (not all the way just the first number) so you do 15x3 and that's 45, then you subtract that from 55, and you get 10, so then you take your denominator (15) and you answer is 3 10/15. But when you simplify it it's 3 and 2/3
Hope I helped sorry it's so long and sorry for any typos it's so long I didn't go back and check
Answer:
1.f(x)=2x-5
i will take the set {-2,-1,0,1,2}
f(-2)=2(-2)-5
=-4-5
=-9
f(-1)=2(-1)-5
=-2-5
=-7
f(0)=2(0)-5
=-5
f(1)=2(1)-5
=-3
f(2)=2(2)-5
=-1
so the coordinates of the function is {-9,-7,-5,-3,-1}
2.f(x)=-3x+6
i will the take the set {-2,-1,0,1,2} too
f(-2)=-3(-2)+6=6+6=12
f(-1)=-3(-1)+6=3+6=9
f(0)=-3(0)+6=6
f(1)=-3(1)+6=-3+6=3
f(2)=-3(2)+6=-6+6=0
{12,9,6,3,0}
3.f(x)=2/3.x+4
{-2,-1,0,1,2}
f(-2)=2/3(-2)+4=-4/3+4=(-4+12)/3=8/3
f(-1)=2/3(-1)+4=-2/3+4=(-2+12)/3=10/3
f(0)=2/3(0)+4=4
f(1)=2/3(1)+4=2/3+4=(2+12)/3=14/3
f(2)=2/3(2)+4=4/3+4=(4+12)/3=16/3
{8/3,10/3,4,14/3,16/3}
you're can graph those coordinates
actually you can take other coordinates...
CMIIW
,
Answer:
301.44 mm
Step-by-step explanation:
Diameter of the inside ring of a packing tape = 80 mm
The tape surrounding the ring is 8 mm beyond the ring which means the diameter of the ring including the surrounding will be:
80 + 8 + 8 = 96 mm
and its radius will be = 96/2 = 48.
To find the distance around the outer edge of the tape, we need to calculate the circumference for the ring with the tape on it.
<em>Circumference = 
</em>
Circumference = 2 x 3.14 x 48 = 301.44 mm
Therefore, the distance around the outer edge of the tape is 301.44 mm.
