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3 years ago

Solve using substitution-2x-y=-9. 5x-2y=18

1 answer:

5 0

Equation 1) -2x-y=-9
Equation 2) 5x-2y=18

Multiply all of equation 1 by 2, so that both equations have (2y) in them.

1) 2(-2x-y=-9)

Simplify.

1) -4x-2y=-18
2) 5x-2y=18

Subtract equations with each other.

-9x=-36

Divide both sides by -9.

x=4

Plug in 4 for (x) into equation 1.

-2x-y=-9

-2(4)-y=-9

Simplify.

-8-y=-9

Add 8 to both sides.

-y=-1

Divide both sides by -1.

y=1

(4, 1)

~Hope I helped!~

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Which of the following expresses the possible number of positive real solutions for the polynomial equation shown below?

sesenic [268]

Answer: D. Two or zero

Step-by-step explanation: x=-7/6+-1/6√385

There isn't an exact place you could really solve this number, regardless of the way you break it down

5 0

3 years ago

I don’t understand please help!

nikklg [1K]

Answer:

x > 10/3

x will always be greater that 10/3 never the same nor less.

Step-by-step explanation:

3/5x + 1/3 < 4/5x - 1/3

3/5x - 4/5x + 1/3 < -1/3

-1/5x + 1/3 < -1/3

-1/5x < -1/3 - 1/3

-1/5x < -2/3

-1/5x/-1 < -2/3/-1 (Since it is both divided by a negative the sign switches, the sign also switches with multiplication. *Note: it only switches if it is negative!*)

1/5x > 2/3

5 • 1/5x > 2/3 • 5

x > 2/3 • 5/1

x > 10/3

6 0

3 years ago

Data on the blood cholesterol levels of 6 rats give mean = 85, s= 12. A 95% confidence interval for the mean blood cholesterol o

lana [24]

The <em><u>correct answer</u></em> is:

c)75.4 to 94.6

Explanation:

The formula for a confidence interval is:

$\mu \pm z*(\frac{\sigma}{\sqrt{n}})$ ,

where μ is the mean, z is the z-score associated with the level of confidence we want, σ is the standard deviation, and n is the sample size.

Our mean is 85, our standard deviation is 12, our sample size is 6, and since we want 95% confidence, our z-score is 1.96:

$85\pm 1.96(\frac{12}{\sqrt{6}})=85\pm 9.6=85-9.6, 85+9.6=75.4, 94.6$

4 0

3 years ago

Read 2 more answers

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

3 years ago

Which ordered pairs make the equation true?

Nata [24]

(-4,-2)
You just plug in the numbers from the ordered pair, into your equation. Remember, the formula for ordered pairs are (x,y). :)

6 0

3 years ago

Read 2 more answers