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2 years ago

What are the two constants of proportionality for the relationship between distance in kilometers

1 answer:

5 0

We call the ratio between two directly proportional quantities the constant of proportionality. When these two quantities may increase or decrease, the constant proportionality always remains the same

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3. The sum of three consecutive numbers<br> is 42. What are the numbers?

Zanzabum

Answer:

13 + 14 + 15

Step-by-step explanation:

n + (n + 1) + (n +2) =42

n + n + n + 1 + 2 = 42

3n + 3 = 42

(subtract 3 from both sides)

3n = 39

(divide both sides by 3)

n = 13

(solve)

n + (n + 1) + (n +2) =42

13 + (13 + 1) + (13 + 2) = 42

13 + (14) + (15) = 42

13, 14, 15

(Hope this helps, have a nice day!)

3 0

3 years ago

Two times the sum of a number and 8 is 20 what is the number

Andre45 [30]

Answer:

n=2

2(n+8)=20

2n+16=20

-16 -16

2 2

n= 2

There u go love

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

David had $35 to spend at the fair. If the admission to the fair is $4.5 and the rides cost $1.50 each, what is the greatest num

sleet_krkn [62]

Answer:

20 rides

Step-by-step explanation:

35-4.5=30.5

20*1.5=30 and 30 plus 4.5 equals 34.5

5 0

2 years ago

A robot can complete up to 8 tasks in 5/6 hour. each task takes the same amount of time.

storchak [24]

$\bf \begin{array}{ccll} tasks&hours\\ \cline{1-2}&\\ 8&\frac{5}{6}\\\\ 1&h \end{array}\implies \cfrac{8}{1}=\cfrac{~~\frac{5}{6}~~}{h}\implies 8=\cfrac{~~\frac{5}{6}~~}{\frac{h}{1}}\implies 8=\cfrac{5}{6}\cdot \cfrac{1}{h} \\\\\\ 8=\cfrac{5}{6h}\implies 48h=5\implies h=\cfrac{5}{48}\\\\[-0.35em] \rule{31em}{0.25pt}$

$\bf \begin{array}{ccll} tasks&hours\\ \cline{1-2}&\\ 8&\frac{5}{6}\\\\ t&1 \end{array}\implies \cfrac{8}{t}=\cfrac{~~\frac{5}{6}~~}{1}\implies \cfrac{8}{t}=\cfrac{5}{6}\implies 48=5t \\\\\\ \cfrac{48}{5}=t\implies 9\frac{3}{5}=t$

7 0

3 years ago

Read 2 more answers

Which is the answer. A B C D?

MakcuM [25]

Answer:

I’m pretty sure it’s B

Step-by-step explanation:

4 0

2 years ago