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LenKa [72]
4 years ago
11

For the following true conditional statement, write the converse. If the converse is also true, combine the statements as a bico

nditional. if x = 3,then x^2 = 9
Mathematics
1 answer:
Licemer1 [7]4 years ago
5 0
Original Statement: If x = 3, then x^2 = 9 (true)

Converse: If x^2 = 9, then x = 3 (false; this is half the story and x = -3 must be included)

Since the converse isn't completely true, we can't combine the two statements to get a biconditional. 
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The sum of two numbers is 125 and their difference is 47 . What are the two numbers
svetoff [14.1K]
X + y = 125
x - y = 47
---------------add
2x = 172
x = 172/2
x = 86

x + y = 125
86 + y = 125
y = 125 - 86
y = 39

ur 2 numbers are 39 and 86
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4 years ago
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Please help!!!!! Thank you :)
MrRa [10]

Answer:

x + 131 + 108 + 107 + 110 = (5 - 2)·180 --> x = 84

3 0
3 years ago
Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
3 years ago
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An animal that has body parts that are arranged around a central point has
sammy [17]
Radically symmetrical
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3 years ago
The piggy bank has 71 dimes and quarters with a
vesna_86 [32]

Answer:

35 Quarters and 36 Dimes

Step-by-step explanation:

d + q = 71

10 + 25 = 1235

10 x 36 = 360

25 x 35 = 875

360 + 875 = 1235

5 0
3 years ago
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