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maxonik [38]
3 years ago
6

Question 3: Add 9 to 5 times n to get 3 is reprented as​

Mathematics
2 answers:
iragen [17]3 years ago
6 0

Answer:

5n + 9 = 3

I hope im right!!

Llana [10]3 years ago
3 0
5n + 9 = 3

because it says add 9 , then you do 5xn which is 5n & you would get 3
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Korolek [52]
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3 years ago
How to graph the line y=4/3x
ankoles [38]

Answer:

make a table of values

Step-by-step explanation:

then plot using those values

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3 years ago
Select all that apply.
Anarel [89]
So the answer your looking for is X=2
5 0
2 years ago
Two runners one averaging 5 miles per hour and the other one averaging 4 miles per hour, start at the same place and run along t
Vikentia [17]

Answer:

The Distance cover by both the Runners is same = 10 miles  

Step-by-step explanation:

According to question ,

The Speed of first runners (S 1) = 5 miles per hour

The speed of second runners (S 2)  = 4 miles per hour

Let The Time taken by First runner (T 1 ) = T hour

But the second runner  arrives half hour after the first runner ,

I.e The Time taken by Second runner (T 2) = ( T + \frac{1}{2} )

Now from Distance = Speed × Time

Since both the runners start from same place and run along the same trail

SO ,Both the Distance cover by both are same , D 1 = D 2

i.e Speed 1 × Time 1 = Speed 2 × Time 2

    5 mph × T            =  4 mph   ×  ( T + \frac{1}{2} )

    5 T =  4 T + ( 4 × \frac{1}{2} )

Or,  5 T - 4 T = 2

∴       T    =  2 hour  ,

Time take by first = T1 = T = 2 hour

Time take by second = T2 = T + \frac{1}{2} = (2  +  \frac{1}{2} )hour = \frac{5}{2}

Now the Distance cover = Speed × Time

              Distance   (D1)          = 5 mph  × 2 hour = 10 mile

And        Distance    (D2)         =  4 mph  × \frac{5}{2} = 10 miles

Hence, As The Distance cover by both the Runners is same = 10 miles  Answer

5 0
3 years ago
If y-3x=9, 7x+y=25, what is the value of x and y?
lianna [129]

\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

4 0
2 years ago
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