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gavmur [86]
3 years ago
9

X + 3y = 2 -x + 2y = 3 Can somebody help me?

Mathematics
1 answer:
Alisiya [41]3 years ago
4 0
X = -1 ; y = 1

x + 3y = 2, isolate x by subtracting 3y from both sides. Now that x is alone (because x = -3y + 2) take -3y + 2 and plug it into the second equation.

-(-3y + 2) + 2y =3 ; multiply what’s in the parenthesis by -1. It now is 3y -2 + 2y =3, add the common variables ( 3y + 2y) and you now have 5y-2=3. Move the 2 to the other side by adding it to both sides, you now have 5y=5, divide both sides by 5 and y=1.

Take any equation and plug y in for 1 to get x.

For example, (in the second equation), -x+2(1)=3, multiply 2 by 1 and you’re left with -x+2=3. Subtract 2 on both sides and you have -x=1, now divide both sides by -1 and x=-1.

You can check your work by plugging in the x and/or y values into the equations
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A collection of dimes and quarters is worth $9.55. If the quarters were dimes and the dimes were quarters, the total value would
kvasek [131]

Number of dimes are 18 and number of quarters are 31

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

A collection of dimes and quarters is worth $9.55

value of 1 dime x number of dimes + value of 1 dime x number of quarters = 9.55

0.10d + 0.25q = 9.55 ---------- eqn 1

If the quarters were dimes and the dimes were quarters, the total value would be 7.60

quarters were dimes means , q = d

dimes were quarters means d = q

0.25d + 0.10q = 7.60 ----- eqn 2

Let us solve eqn 1 and eqn 2 to find "d" and "q"

Multiply eqn 1 by 2.5

0.25d + 0.625q = 23.875 ---- eqn 3

Subtract eqn 2 from eqn 3

0.25d + 0.625q = 23.875

0.25d + 0.10q = 7.60

( - ) ----------------------

0.525q = 16.275

q = 31

Substitute q = 31 in eqn 1

0.10d + 0.25q = 9.55

0.10d + 0.25(31) = 9.55

0.10d + 7.75 = 9.55

0.10d = 1.8

d = 18

Thus dimes are 18 and number of quarters are 31

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lateral surface area of this cylinder is 1584/7 or 226.28...cm square.

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Determine the end behavior for function f(x)=-x^4+5x^3-3
Gnesinka [82]

Answer:

Step-by-step explanation:

The dominant term of this function is x^4.  The graph of x^4 starts in Quadrant II and continues in Quadrant I.

If we have y = -x^4, the graph starts in Quadrant III and continues in Quadrant IV.  This is the end behavior for f(x)=-x^4+5x^3-3.

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3 years ago
the second term of a geometric sequence is 18 and the fourth term is 8 find the common ratio . And find the sum of the first 6 t
777dan777 [17]

Answer:

  • ratio: 2/3
  • sum: 73 8/9

Step-by-step explanation:

The general term of a geometric sequence is ...

  an = a1·r^(n-1)

You have the 2nd and 4th terms, so ...

  a2 = a1·r^(2-1) = a1·r

  a4 = a1·r^(4-1) = a1·r^3

We can find r from the ratio ...

  a4/a2 = (a1·r^3)/(a1·r) = r^2 = 8/18 = 4/9

Then r is ...

  r = √(4/9) = 2/3 . . . . the common ratio

The first term is ...

  a2 = 18 = a1·(2/3)

  a1 = (3/2)·18 = 27

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The sum of the first 6 terms is ...

  Sn = a1·(r^n -1)/(r -1)

  S6 = 27·((2/3)^6 -1)/(2/3 -1)

  S6 = 27·(64/729-1)/(2/3-1) = (27)(665)/243 = 73 8/9

The sum of the first 6 terms is 73 8/9.

_____

<em>Check on the sum</em>

The first 6 terms are ...

  27, 18, 12, 8, 5 1/3, 3 5/9

Their sum is 73 8/9, as above.

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Sliva [168]

Answer:

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Step-by-step explanation:

(a) The linear approximation of g(x) at x=b will be ...

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Using the given relations, this is ...

  g'(3) = 3² +7 = 16

  g(x) ≈ 16(x -3) -1

Then the points of interest are ...

  g(2.95) ≈ 16(2.95 -3) -1 = -1.8

  g(3.05) ≈ 16(3.05 -3) -1 = -0.2

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(b) At x=3, the slope of the curve is increasing, so the tangent lies below the curve. The estimates are too small. (Matches description A.)

4 0
3 years ago
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