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ArbitrLikvidat [17]
2 years ago
15

Y=3x+62x+4y=−4 What is the value of y in the system?

Mathematics
1 answer:
Marrrta [24]2 years ago
6 0

9514 1404 393

Answer:

  y = 0

Step-by-step explanation:

We assume you want to find y such that ...

  • y = 3x +6
  • 2x +4y = -4

Dividing the second equation by 2, we can write an expression for x.

  x +2y = -2

  x = -2y -2

Substituting this into the first equation gives ...

  y = 3(-2y -2) +6

  y = -6y -6 +6 . . . eliminate parentheses

  7y = 0 . . . . . . . . . add 6y

  y = 0 . . . . . . . . . . divide by 7

The value of y in the system of equations is 0.

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The Frosty Ice-Cream Shop sells sundaes for $3 and banana splits for $5. Yesterday the amount of banana splits sold was 14 less
Vladimir [108]

Answer:

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6 0
2 years ago
Find the sum of the first 6 terms of the following series, to the nearest integer.
natka813 [3]

Answer:

1008

Step-by-step explanation:

geometric sequence:

a=16

r=32/16=2

Sum of GS:

s=a(1-r^n)/(1-r)

S(6)=16(1-2^6)/(1-2)

=1008

8 0
3 years ago
Sita Rahim drove due north for 3 hours at 46 miles per hour. From the
Phantasy [73]

Answer:

224 miles

Step-by-step explanation:

What you would do is take the 3 hours times the 46 mph to get 138 miles.

3*46=138

Next, multiply the 2 hours by the 43 mph to get 86 miles.

2*43=86

Lastly, you're going to add the two.

138+86=224 miles.

6 0
3 years ago
Which function in vertex form is equivalent to f(x) = x^2 + 6x + 3?
OlgaM077 [116]
f(x)=x^2+6x+3

Complete the square:

x^2+6x+a^2\implies a=\dfrac{6x}{2x}=3\implies x^2+6x+3^3

And we have:

f(x)=x^2+6x+3\\\\f(x)=x^2+6x+3^2-3^2+3\\\\f(x)=(x+3)^2-9+3\\\\\boxed{f(x)=(x+3)^2-6}

Answer B.

3 0
3 years ago
PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM
Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer:  {x = 1 , y=1, z=0

6 0
2 years ago
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