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gladu [14]
3 years ago
10

Given: a||b, m∠3=63° Find: m∠6, m∠7

Mathematics
2 answers:
jonny [76]3 years ago
6 0
We need an image of this
FromTheMoon [43]3 years ago
5 0

Answer:

Where's the figure,my friend?

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Prove that (sec theta-cos theta) ( cot theta+tan theta) = tan theta ×sec theta​
meriva

Given:

(\sec \theta \cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta

To prove:

The given statement.

Proof:

We have,

(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta

Taking LHS, we get

LHS=(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)

LHS=(\dfrac{1}{\cos \theta }-\cos \theta)(\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta})

LHS=(\dfrac{1-\cos^2 \theta }{\cos \theta })(\dfrac{\cos^2 \theta+\sin^2 \theta}{\sin \theta\cos \theta})

LHS=(\dfrac{\sin^2 \theta }{\cos \theta })(\dfrac{1}{\sin \theta\cos \theta})          [\because \cos^2 \theta+\sin^2 \theta=1]

LHS=(\dfrac{\sin \theta }{\cos \theta })(\dfrac{1}{\cos \theta})

LHS=\tan \theta \sec \theta

LHS=RHS

Hence proved.

3 0
3 years ago
(8x) 2/3 radical form
Tju [1.3M]
16 3 x the took a quiz
3 0
3 years ago
If y= (2x^2+x)^3 +(x-1) /(1-x). find dy dx​
Mrrafil [7]

Answer:

Step-by-step explanation:

y=(2x²+x)³+(x-1)/(1-x)

=(2x²+x)³+(x-1)/-(x-1)

=(2x²+x)³-1

dy/dx=3(2x²+x)(4x+1)-0

=3x(2x+1)(4x+1)

7 0
3 years ago
The Table Shows Ryan's Total Skiing Distance At Different Time Intervals During His Cross-Country Ski Outing. What Was The Avera
creativ13 [48]
57-15/ 8-2
the answer would be 7

8 0
3 years ago
Matt can walk 500 m from his home on King Street, and then 900m along Queen Street to go to his school. He can also walk across
brilliants [131]

Answer:

He walks 1030 meters if he walks across the park.

Step-by-step explanation:

In order to solve this question we will need to know that c^{2} = a^{2} + b^{2} (were c is the length of the hypotenuse (the diagonal line) of a right angle triangle, and a and b are the legs (the sides that form a right angle). So them means that.....

Let "c" be the distance he will need to walk if he was going to go through the park

Let "a" be the distance he walks on King Street

Let "b " be the distance he walks on Queen Street, then.....

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

c = \sqrt{a^{2} + b^{2} }

(Now plug in the values of a and b and get.....)

c = \sqrt{500^{2} + 900^{2}  }

c = \sqrt{1060000 }

c = 1029.563014............

So I would assume that you have to round to the nearest meter. And as a result we get 1030 meter.

7 0
3 years ago
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