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Juli2301 [7.4K]

3 years ago

5

X²-gx Which shows all the critical points for the inequality 3-5 <0?

1 answer:

den301095 [7]3 years ago

5 0

Answer: will be = -2<0

Step-by-step explanation:x2 -gx=3-5<0=-2<0

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Which expression is equivalent to sq root 36/25

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Is this multiple choice ? If so please list the answers

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How do you know that the systems of equations are consistent? How do you know that they are independent?

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Answer:

If a system has at least one solution, it is said to be consistent. If a consistent system has exactly one solution, it is independent.

Step-by-step explanation:

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3 years ago

What <br> is (81m^6)^1/2 simplified

ANEK [815]

(81m^6)^1/2 = (3^4*1/2) (m^6*1/2) = 3²m³

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3 years ago

Lines k and n intersect on the y-axis

avanturin [10]

a) The equation of line k is:

$y = -\frac{202}{167}x + \frac{598}{167}$

b) The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

The equation of a line, in <u>slope-intercept formula</u>, is given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change.
b is the y-intercept, which is the value of y when x = 0.

Item a:

Line k intersects line m with an angle of 109º, thus:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

In which $m_1$ and $m_2$ are the slopes of <u>k and m.</u>

Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

$m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}$

The tangent of 109 degrees is $\tan{109^{\circ}} = -\frac{29}{10}$
Thus, the slope of line m is found solving the following equation:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

$-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}$

$m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2$

$m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}$

$\frac{167m_2}{80} = \frac{-202}{80}$

$m_2 = -\frac{202}{167}$

Thus:

$y = -\frac{202}{167}x + b$

It goes through point (-2,6), that is, when $x = -2, y = 6$ , and this is used to find b.

$y = -\frac{202}{167}x + b$

$6 = -\frac{202}{167}(-2) + b$

$b = 6 - \frac{404}{167}$

$b = \frac{6(167)-404}{167}$

$b = \frac{598}{167}$

Thus. the equation of line k, in slope-intercept formula, is:

$y = -\frac{202}{167}x + \frac{598}{167}$

Item b:

Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

$-\frac{202}{167}m = -1$

$m = \frac{167}{202}$

Then

$y = \frac{167}{202}x + b$

Also goes through point (-2,6), thus:

$6 = \frac{167}{202}(-2) + b$

$b = \frac{(2)167 + 202(6)}{202}$

$b = \frac{1546}{202}$

The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

A similar problem is given at brainly.com/question/16302622

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2 years ago

To determine the inverse of a point, you?

oksano4ka [1.4K]

So if you're asked to graph a function and its inverse, all you have to do is graph the function and then switch all x and y values in each point to graph the inverse. Just look at all those values switching places from the f(x) function to its inverse g(x) (and back again), reflected over the line y = x.

8 0

3 years ago