PLEASE HELP URGENT
Maggie is walking her dog. She walks 2/3 mile in 7/10 of an hour. At this same rate, what distances can she walk? Select all that apply.
5/7 mile in 3/4 of an hour
7/10 mike in 2/3 of an hour
10/3 miles in 1 1/2 hours
32/21 miles in 1 3/5 hours
5/3 mikes in 1 9/10 hours
Answers
We want to find the mile per hour rate (mi/hr).
That is 2/3 divided by 7/10
Which is 2/3 times 10/7
Numerator is 20
Denominator is 21
So 20/21 miles per hour. .95238009…
Convert each answer (flipping the second fraction since it is division)
5/7 times 4/3 is 20/21
7/10 times 3/2 is 21/20
10/3 divided by 1.5 hours which is 3/2.
So 10/3 times 2/3 is 20/9
32/21 divided by 1 3/5 hours which is 5/5 plus 3/5 (add to get 8/5)
So 32/21 times 5/8 is 160/168
5/3 times 10/9 is 50/27
In order of answers
20/21
21/20
20/9
160/168
50/27
Original was 20/21 which is less than 1 (21/21)
So yes to 20/21
So no to 21/20, 20/9, 50/27
To see if 160/168 works simplify.
160/168 gives 80/84 gives 40/42 gives 20/21 when dividing by 2.
So 160/168 works.
So yes to
5/7 mile in 3/4 of an hour
32/21 miles in 1 3/5 hours
So no to
7/10 mile in 2/3 of an hour
10/3 miles in 1 1/2 hours
5/3 miles in 1 9/10 hours
Please mark as the best!
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Hey there!
2(3a - 5b) + 4(a + 2b)
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Good luck on your assignment & enjoy your day!
~Amphitrite1040:)
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Answers
Answer:
- 98Step-by-step explanation:
Given
[tex]a_{21}=-32[/tex][tex]a_{50}=-119[/tex]Find [tex]a_{43}[/tex]
We know that
[tex]a_n=a_1+(n-1)d[/tex]Appy this to the two known terms and find the first term and the common difference
[tex]a_1+20d=-32[/tex] ⇒ [tex]a_1=-20d-32[/tex] [tex]a_1+49d=-119[/tex] ⇒ [tex]a_1=-49d-119[/tex]Find the coomon difference
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a₄₃ = 28 + 42(- 3) = 28 - 126 = - 98Answer:
-98
Step-by-step explanation:
General form of an arithmetic sequence:
[tex]a_n=a+(n-1)d[/tex]
where:
[tex]a_n[/tex] is the nth terma is the first termd is the common difference between termsGiven:
[tex]a_{21}=-32[/tex][tex]a_{50}=-119[/tex]To find the 43rd term we need to construct an equation for the nth term. To do this, find the first term and the common difference by substituting the given information into the general formula to create a system of equations that can be solved.
[tex]\begin{aligned} \underline{\textsf{Equation 1}}} \\ a_{21} & =-32 \\ \implies a+(20-1)d & = -32 \\a+20d & =-32 \end{aligned}[/tex]
[tex]\begin{aligned} \underline{\textsf{Equation 2}}} \\ a_{50} & =-119 \\ \implies a+(50-1)d & = -119 \\a+49d & =-119 \end{aligned}[/tex]
Subtract Equation 1 from Equation 2 to eliminate a:
[tex]\begin{array}{r l}a+49d & = -119\\- \quad a+20d & = -32\\\cline{1-2}29d & = -87\end{aligned}[/tex]
Solve for d:
[tex]\implies d=\dfrac{-87}{29}=-3[/tex]
Substitute the found value of d into one of the equations and solve for a:
[tex]\implies a + 20(-3)=-32[/tex]
[tex]\implies a=28[/tex]
Substitute the found values of a and d into the general formula to create an equation for the nth term:
[tex]\implies a_n=28+(n-1)(-3)[/tex]
[tex]\implies a_n=31-3n[/tex]
Finally, substitute n = 43 into the found formula to find the 43rd term of the sequence:
[tex]\implies a_{43}=31-3(43)=-98[/tex]
Learn more about arithmetic sequences here:
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