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Ilya [14]
3 years ago
11

Ive been struggling with this problem can someone help. Thank you

Mathematics
1 answer:
borishaifa [10]3 years ago
3 0

Answer:

70 degrees

Step-by-step explanation:

Angles 130 degrees and TUV are supplementary angles.  Thus, the latter is (180 - 120), or 60, degrees.  Now the sum of all three interior angles of this triangle must be 180 degrees:  (60 + 50 + >TUV) = 180, so >TUV is 70 degrees (answer C).

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Taika has 320.00 in the bank. He spends 5% on a gift for his sister. How much did he spend?
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He spent 16 dollars on his sisters gift.

Step-by-step explanation:

You can do 320 x .05 which will equal 16.

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Explain why both Jenna and Mia arrived at the same answer and the advantage of one method over the other.
Lynna [10]

Complete question :

Jenna’s method: Mia’s method: 5(30 + 4) 5(30 + 4) (5)(30) + (5)(4) 5(34) = 170 150 + 20 = 170 Explain why both Jenna and Mia arrived at the same answer and the advantage of one method over the other.

Answer:

Kindly check explanation

Step-by-step explanation:

Jenna's method :

5(30 + 4) = 5(34) = 170

Mia’s method :

5(30 + 4) :

5(30) + 5(4)

150 + 20

= 170

With both methods, we will arrive at the same answer as justified in the workings above. However, Jenna's method gives a very simple approach and less complicated arithmetic by using the distributive property to overrode some multiplication process which Mia went through in her own calculation. Therefore, Jenna's method is less complicated and requires two simple steps at obtaining the final result.

4 0
3 years ago
Which of the following are solutions to 2tanx/1-tan^2x=sqrt 3?
Tanya [424]

Answer:

A, D

Step-by-step explanation:

(\frac{2tanx}{1-tan^{2}x}) =\sqrt{3} (1)

Assume that tan x = y. Replace this into the equation (1), we have:

+) \frac{2y}{1-y^{2} } =\sqrt{3}

=> 2y = \sqrt{3}  . (1 -y^{2} ) =- \sqrt{3} .y^{2}  + \sqrt{3}

=> \sqrt{3} .y^{2}  + 2y - \sqrt{3} = 0

=> \sqrt{3} . y^{2} - 1.y + 3.y - \sqrt{3} =0

=> (\sqrt{3} . y.y - 1.y) + (\sqrt{3}.\sqrt{3}  .y - \sqrt{3}) =0

=> y.(\sqrt{3} y - 1) + \sqrt{3} (\sqrt{3} y - 1) = 0

=> (y + \sqrt{3} ).(\sqrt{3} .y -1)=0

=> y + \sqrt{3} = 0 or \sqrt{3} y -1 =0

If y + \sqrt{3} = 0

=> y = - √3

=> tan x = - √3

=> x = \frac{2\pi }{3} + n.\pi with n is an integral

If \sqrt{3} y -1 =0

=> y = 1/√3

=> tan x = 1/√3

=> x = \frac{\pi }{6} + n.\pi n is an integral

So that A, D are answers.

3 0
3 years ago
Read 2 more answers
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