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Dmitriy789 [7]
3 years ago
15

(False Converses) Write down the converse of each conditional statement about the integers. Then give a reason why the converse

is false.
If x is odd and y is odd, then x + y is even.
Mathematics
1 answer:
Mumz [18]3 years ago
8 0

Answer:

Converse: “If x+y is even, then x is odd and y is odd”

FALSE

Step-by-step explanation:

The converse of

“If x is odd and y is odd, then x+y is even”

is

“If x+y is even, then x is odd and y is odd”

The converse is false.

Counter-example:

Take x = 2, y = 4

Then x+y = 6 which is even, but neither x nor y are odd.

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A triangle has sides of lengths 9, 7, and 12. Is it a right triangle? Explain.
Brut [27]

Answer:

Yes based on the numbers .

Step-by-step explanation:

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3 years ago
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Which of the following is the inverse of y=6^x? y=log_(6)x, y=log_(x)6, y=log_((1)/(6))x, y=log_(6)6x
kolezko [41]

Answer:

y=log_(6)x

Step-by-step explanation:

From the definition of logarithm:

x = log_b y <---> y=b^x

In our case, we have

y=6^x

which means that b=6, therefore the inverse function is

x=log_6 y

which can be rewritten by switching x and y:

y=log_6 x

4 0
3 years ago
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Use the drop-down menus to complete each equation so the statement about its solution is true No Solutions 2T +5 2r 3r T One Sol
Mandarinka [93]

Answer:

asdasdasdasdasdasdsdsad

Step-by-step explanation:

asdasdasdasdasdsdasdasdasd

5 0
3 years ago
If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn
mash [69]

Let a=\tan A and b=\sin A. Then

m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab

\implies m^2-n^2=4\tan A\sin A

and

mn=(a+b)(a-b)=a^2-b^2

\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}

The expression under the square root can be rewritten as

\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)

Recall that

\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A

so that

\tan^2A-\sin^2A=\sin^2A\tan^2A

and assuming \sin A>0 and \tan A>0, we end up with

4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A

so that

m^2-n^2=4\sqrt{mn}

as required.

5 0
3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
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