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Talja [164]
3 years ago
6

2.) a. If you rotate an object 180 degrees clockwise or counterclockwise, using the same center, is the image the same in both c

ases? Explain.. b. Answer part (a) if you rotate the object 360 degrees..
Mathematics
1 answer:
pentagon [3]3 years ago
4 0
An object rotated by 180 degrees in a clockwise and counterclockwise can have axis reversed. Hence the image is not the same anymore. However when it is 360 degrees reversed, the image is brought back to its original axis, hence the image is the same again,
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(2.3 x 10 to the power of 4) times ( 1.5x10 to the power of minus 3) in standard form
Eduardwww [97]

Answer:

3.45\times10

Step-by-step explanation:

(2.3  \times 10^{4}) \times ( 1.5 \times10^{-3} )= 2.3\times1.5\times10^{4}\times10^{-3}

                                        = 3.45\times  10^{((4)+(-3))}

                                        =3.45\times10

6 0
3 years ago
Can smby plz help me ?!
NARA [144]
Omgggg an iready diagnostic..
6 0
3 years ago
Jane and Lee had dinner at The Palace The bill totaled $20 30 with tax The service
kirza4 [7]

Answer:

$3.45

Step-by-step explanation:

$20.30 / 10  = $2.30 = 10%

$2.30 / 2  = $1.15 = 5%

$2.30 + $1.15 = $3.45 = 15%

3 0
3 years ago
Which is the value of this expression when p=-2 and q=-1?
saul85 [17]

Answer:

D. 4

Step-by-step explanation:

[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4

8 0
3 years ago
If a = 7 and b = 11, what is the measure of ∠B? (round to the nearest tenth of a degree) A) 32.5° B) 39.2° C) 50.5° D) 57.5°
Ilia_Sergeevich [38]

Answer:

D) 57.5°

Step-by-step explanation:

As the question is not complete. So, let's suppose it is a right angle triangle then, we can apply Pythagoras theorem to calculate the hypotenuse or the third side.

Pythagoras Theorem = c^{2} = a^{2} + b^{2}

a = 7 and b = 11

a^{2} = 49

b^{2} = 121

Plugging in the values, we will get:

c^{2} = 49 + 121

c^{2} = 170

c = \sqrt{170}

To calculate the unknown angle B, we can use law of sine.

Law of sine = \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

So,

\frac{c}{sinC} = \frac{b}{sinB}

\frac{\sqrt{170} }{sin90} =  \frac{11}{sinB}

Sin90 = 1

sinB = \frac{11}{\sqrt{170} }

B = sin^{-1} (\frac{11}{\sqrt{170} })

B = 57.5°

8 0
3 years ago
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