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Phantasy [73]
4 years ago
7

What is 110y+10=340 for algebra grade 7

Mathematics
2 answers:
algol [13]4 years ago
7 0
<span> 110y+10=340

Subtract 10 from both sides 

110y=330 

Divide by 110 to get variable on its own 

</span>110y/110=330/110

y=3
natima [27]4 years ago
5 0
We have to multiply by 'y' until we get the value of it

110(y)+10-(340)= ?

Look for your "like terms"
110(y) - 330

Take off 110 make it equal to 0
110 = 0

We have one solution in this problem!

Add the number 3 to your sides (if you want to make it easier )
y - 3 = 0
+ 3 3
---- ---
3y ➡ 3

3 + y = 3y ; 3 + 0 = 3

y = 3
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One positive integer is 5 times another positive integer and their product is 320. What are the positive integers?​
seropon [69]

Answer:

The integers are (x, y) = (40, 8).

Step-by-step explanation:

x = 5y

xy = 320

Substitute the first equation into the second equation.

(5y)(y) = 320

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The integers are (x, y) = (40, 8).

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3 years ago
When they say solve the product in the lowest terms what do they mean?
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8 0
3 years ago
The average (arithmetic mean) of y numbers is x. If 30 is added to the set of numbers, then the average will be x − 5. What is t
Alenkasestr [34]

Answer:

y = \frac{1}{5}(x^{2}-35x)

Step-by-step explanation:

Let the total numbers are n.

If the average of y numbers is x then we can form an equation

\frac{y}{n}=x

⇒ \frac{n}{y}=\frac{1}{x}

⇒ n = \frac{y}{x} --------(1)

Now 30 is added to the set of numbers then average becomes (x - 5)

\frac{y+30}{n+1}=(x-5)

⇒ \frac{(n+1)}{(y+30)}=\frac{1}{(x-5)}

⇒ (n + 1) = \frac{y+30}{x-5}

⇒ n = \frac{y+30}{x-5} - 1 ----- (2)

Now we equate the values of n from equation 1 and 2

\frac{y}{x} = \frac{y+30}{x-5} - 1

y(x - 5) = x(y + 30) - x(x - 5)  [ By cross multiplication ]

xy - 5y = xy + 30x - x² + 5x

xy - xy - 5y = 35x - x²

-5y = 35x - x²

x² - 35x = 5y

y = \frac{1}{5}(x^{2}-35x)

6 0
3 years ago
Read 2 more answers
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