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marta [7]
3 years ago
14

19. Simplify 3√ 2 – √ 2 . A. 4√ 2 B. 2√ 2 C. √ 2 D. 3√ 2

Mathematics
2 answers:
Triss [41]3 years ago
6 0

Answer:

The correct answer is option B.  2√ 2

Step-by-step explanation:

It is given that, 3√ 2 – √ 2

<u>To find the correct option</u>

3√ 2 – √ 2  can be written as,

3√ 2 – √ 2  = √ 2 (3 - 1)   taking √ 2  as common

 = √ 2  * 2

 = 2√ 2

Therefore the correct answer is 2√ 2 .

The correct option is option b.  2√ 2

podryga [215]3 years ago
3 0

Answer: OPTION B

Step-by-step explanation:

You need to remember that, to subtract radicals, the indices and the radicands must be the same.

Given the expression 3\sqrt{2}-\sqrt{2}, you can identify that both have index 2 and the radicands are 2 (The numbers that are inside of radicals), therefore, you can conclude that the subtraction can be made.

Then, you must subtract the terms in front of the radicals. Therefore, you get:

3\sqrt{2}-\sqrt{2}=2\sqrt{2}

You can observe that this matches with the option B.

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ziro4ka [17]
Answer: if Gregory draws the segment with endpoints A and A’, then the midpoint will lie on the line of reflection.

Explanation:

Given that a triangle ABC is reflected in triangle A'B'C'

Here reflection is done on a line

If you imagine the line as a mirror then ABC will have image on the mirror line as A'B'C'

Recall that in a mirror the object and image would be equidistant from the mirror and also the line joining the image and object would be perpendicular to the mirror

But note that corresponding images will only be perpendicular bisector to the line

So A and A' only will be corresponding so AA' will have mid point on line

Option 1 is right
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3 years ago
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\huge {\mathfrak{Answer :  }}

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\boxed{\cos( \theta)  =  \dfrac{base}{hypotenuse}}

so,

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now,

\boxed{ \tan( \theta) =  \frac{perpendcular}{base}  }

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3 years ago
Help me with my math quiz from friday 4/1 <br> SHOW STEPS
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Hie!! I will be solving the questions in the sequence of the attachments.

_______________________________________

4. As line SU and TV are intersecting each other at O.

A.T.Q, ∠UOV = (7x - 4)°

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Then, ∠UOV = ∠SOT {Vertically Opposite Angles}

\rightarrow\sf{7x - 4=87}

\rightarrow\sf{7x =87+4}

\rightarrow\sf{7x =91}

\rightarrow\sf{x =\frac{91}{7}}

\rightarrow\boxed{\bf{x =13}}

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5. radius = 17ft

circumference of circle = 2πr

\rightarrow\sf{circumference=2\times\frac{22}{7}\times17}

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3. ∠1 and ∠2 - add to 180 (linear pair)

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_______________________________________

2. 160° + (6x - 16)° = 180° (linear pair)

\rightarrow\sf{6x - 16 = 180-160}

\rightarrow\sf{6x - 16 = 20}

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\rightarrow\sf{6x =36}

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Find the area of the surface. the part of the surface z = xy that lies within the cylinder x2 + y2 = 81
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