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telo118 [61]
3 years ago
13

Solve the Equation for y . 9x +5y = -2 ​

Mathematics
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

\large\boxed{y=\dfrac{-9x-2}{5}}

Step-by-step explanation:

9x+5y=-2\qquad\text{subtract}\ 9x\ \text{from both sides}\\\\5y=-9x-2\qquad\text{divide both sides by 5}\\\\y=\dfrac{-9x-2}{5}

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Simplify x^3y^4/3y^4
yan [13]
The given expression is:
(x^3 * y^4) / (3y^4)

We can notice that the term y^4 is a common term found in both the numerator and the denominator of the given expression, therefore, we can cancel this term from the both numerator and denominator (as if you divided both numerator and denominator by y^4).

Doing this, we will have the simplified form of the expression as follows:
(x^3) / (3)
4 0
3 years ago
Read 2 more answers
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oksian1 [2.3K]

Given :-

  • a² - 2a - b² = 0
  • 2b + 2ab = 0

To find :-

  • Value of a and b .

Solution :-

<u>Taking</u><u> </u><u>second</u><u> </u><u>equation</u><u>:</u><u>-</u>

  • 2b + 2ab = 0
  • 2b ( 1 + a ) = 0
  • 2b = 0 or (1+a) = 0
  • b = 0 , a = -1

<u>Substitute</u><u> </u><u>in </u><u>first </u><u>equation</u><u> </u><u>:</u><u>-</u><u> </u>

  • a² - 2a - b² = 0

<u>When </u><u>b </u><u>=</u><u> </u><u>0</u><u> </u><u>,</u>

  • a² - 2a - 0² = 0
  • a² - a = 0
  • a( a -1) =0
  • a = 0 , 1

<u>When </u><u>a </u><u>=</u><u> </u><u>-</u><u>1</u><u> </u><u>,</u>

  • (-1)² - 2*(-1) - b² = 0
  • 1 + 2 - b² = 0
  • b² = 3
  • b = ±√3

<u>Answer </u><u>:</u><u>-</u><u> </u>

  • a = 0,1 ; b = 0
  • a = -1 , b = ±√3
8 0
3 years ago
Please help!!!will get brainliest!
il63 [147K]
Turn the information into coordinate points.

Point = (time, population)

Point 1 = (1985, 45000)
Point 2 = (2004 , 26000)

Find the slope between these points using the formula

( Slope)—> m = (y2 - y1) / (x2 - x1)
26000-45000/2004-1985=-19000/19= -1000
Average rate of change is decrease of 1000 sea lion per year

I hope that helped
6 0
3 years ago
A clock was reading the time accurately on Friday at noon. On Monday at 6pm the clock was running late by 468 seconds. On averag
Setler [38]

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × \frac{1}{2} = 24

Saturday = 48

Sunday = 48

Monday = 48 × \frac{3}{4} = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds  

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

To learn more about clock problems visit:

brainly.com/question/27122093.

#SPJ1

3 0
2 years ago
PLZZZZZZZZ HELP MEEEEE
lukranit [14]
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