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3 years ago

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What is the converse of the statement? If a point lies in Quadrant III, then its coordinates are both negative. If the coordinat

es of a point are both negative, then the point is in Quadrant III. If the coordinates of a point are both negative, then the point is not in Quadrant I. If a point is in Quadrant I, then its coordinates are both positive. If the coordinates of a point are both negative positive, then the point is not in Quadrant II.

2 answers:

Vladimir [108]3 years ago

6 0

<h2>Answer:</h2>

The converse statement is:

If the coordinates of a point are both negative, then the point is in Quadrant III.

<h2>Step-by-step explanation:</h2>

We know that for any conditional statement of the type:

If p then q i.e. p → q

where p is the hypothesis and q is the conclusion.

The converse of the statement is given by:

If q then p i.e. q → p.

We are given a statement as:

If a point lies in Quadrant III, then its coordinates are both negative.

i.e. Here p=Point lie in Quadrant III

and q= Coordinates are both negative.

Hence, the converse statement will be:

If the coordinates of a point are both negative, then the point is in Quadrant III.

Dovator [93]3 years ago

5 0

<span>If the coordinates of a point are both negative, then the point is in Quadrant III.

</span>

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Valentin [98]

Answer:

slope is undefined

ex. a straight vertical line like x = 2 would have an undefined slope

Step-by-step explanation:

formula is

y2-y1 / x2-x1

-13 + 9 / -5 + 5

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2 years ago

What property is -3(m+8)=-3m-24

Damm [24]

Answer:

All real numbers are solutions.

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−3(m+8)=−3m−24

(−3)(m)+(−3)(8)=−3m+−24(Distribute)

−3m+−24=−3m+−24

−3m−24=−3m−24

Step 2: Add 3m to both sides.

−3m−24+3m=−3m−24+3m

−24=−24

Step 3: Add 24 to both sides.

−24+24=−24+24

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Answer:

All real numbers are solutions.

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2 years ago

2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.

likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

$y = ^+_- \sqrt{-x^2+4x+4} -5$

Thus

$f(x) = \sqrt{-x^2+4x+4} - 5$

$g(x) = -\sqrt{-x^2+4x+4} - 5$

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

$\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0$

The quadratic has roots

$\frac{52 ^+_-\sqrt{1664}}{26}$

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

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2 years ago

Someone please help me ??

KATRIN_1 [288]

Answer:

A filled in circle indicates "or equal to" in a graphing situation. If it's open it means anything less than or greater then that number depending on whatever your graph looks like.

Hopefully this helped!

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3 years ago

Read 2 more answers

Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert

Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

$s=ut+\frac{1}{2} at^{2}$

Here, s = distance of the balloon above the ground = - 80 feet

u = intital velocity = 16 feet per second

a = acceleration of the object = -32 feet per second

t = required time

So, $s=ut+\frac{1}{2} at^{2}$

$-80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})$

$-80=16t-16 t^{2}$

$16 t^{2} -16t -80 =0$

$t^{2} -t -5 =0$

Now, we will use the quadratic D formula for finding the value of t, i.e;

$t = \frac{-b\pm \sqrt{D } }{2a}$

Here, a = 1, b = -1, and c = -5

Also, D = $b^{2} -4ac$ = $(-1)^{2} -(4 \times 1 \times -5)$ = 21

So, $t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}$

$t = \frac{1\pm \sqrt{21 } }{2}$

We will neglect the negative value of t as time can't be negative, so;

$t = \frac{1+ \sqrt{21 } }{2}$ = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

$v=u+at$

Here, v = final velocity

So, $v=16+(-32 \times 2.79)$

v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

8 0

3 years ago