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Assoli18 [71]
3 years ago
10

The sum of y and 2 is equal to 5 less than x

Mathematics
2 answers:
ArbitrLikvidat [17]3 years ago
5 0
 the sum of y and 2 means y+2, is equal to means = and 5 less than x is x-5 so that y + 2 = x-5
ser-zykov [4K]3 years ago
4 0
Y+2=x-5 is the answer
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Tanikwa started a new exercise program using the indoor track. In her first workout she ran 8 laps for every 4 laps that she wal
Elina [12.6K]

Answer:

The correct answer would be 2:1

Step-by-step explanation:

In order to find this, start with the ratio as it is written and sub in the values. Then simplify to the lowest possible terms.

Ran:Walker

8:4

2:1

6 0
3 years ago
Given the system of equations presented here:
Nata [24]
2x + 4y = 14
4x + y = 20......multiply by -4
----------------
2x + 4y = 14
-16x - 4y = -80 (result of multiplying by -4)
---------------add
-14x = -66....as u can see, ur y's cancel out

so ur answer is : 1st answer choice <==

** and just so u know, u could have multiplied the 1st equation by -2, and it would have cancelled out ur x's
6 0
3 years ago
Pls help!! will give all my points
sergejj [24]

when x=-2

-10(-2)+y=4

y=4-20

y=-16

when x=-1

-10(-1)+y=4

y=-6

when x=0

-10(0)+y=4

y=4

when x=1

-10(1)+y=4

y=14

when x=2

-10(2)+y=4

y=24

8 0
3 years ago
Read 2 more answers
Please help i’ll give brainliest
otez555 [7]

Answer:

median

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
FIND THE SUM OF 20 TERMS OF THE ARITHMETIC SERIES IN WHICH 3rd TERM IS 7 AND 7th TERM IS 2 MORE THAN THREE TIMES ITS 3rd TERM
algol13

Answer:

740

Step-by-step explanation:

The n th term of an arithmetic series is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 7 and a₇ = (3 × 7) + 2 = 21 + 2 = 23 , then

a₁ + 2d = 7 → (1)

a₁ + 6d = 23 → (2)

Subtract (1) from (2) term by term

4d = 16 ( divide both sides by 4 )

d = 4

Substitute d = 4 into (1)

a₁ + 2(4) = 7

a₁ + 8 = 7 ( subtract 8 from both sides )

a₁ = - 1

The sum to n terms of an arithmetic series is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ] , thus

S_{20} = \frac{20}{2} [ (2 × - 1) + (19 × 4) ]

     = 10(- 2 + 76) = 10 × 74 = 740

8 0
3 years ago
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