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3 years ago

Just do part B plzz for 12 points i mark brainliest please help me!!

Mathematics

2 answers:

Pani-rosa [81]3 years ago

7 0

Answer:

1/8 i think

Step-by-step explanation:

nignag [31]3 years ago

4 0

Answer:

1/5 or above 5

Step-by-step explanation:

bc 1/5 is smaller than 1/4

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Find the arc length of the partial circle. <br>length is 1.<br>**See attached photo**<br>

KatRina [158]

Arc length of the quarter circle is 1.57 units.

Solution:

Radius of the quarter circle = 1

Center angle (θ) = 90°

To find the arc length of the quarter circle:

$$\text{Arc length}=2 \pi r\left(\frac{\theta}{360^\circ}\right)$

$$=2 \times 3.14 \times 1\left(\frac{90^\circ}{360^\circ}\right)$

$$=2 \times 3.14 \times 1\left(\frac{1}{4}\right)$

Arc length = 1.57 units

Arc length of the quarter circle is 1.57 units.

5 0

3 years ago

Convert 7-6ths into a mixed number

AlexFokin [52]

11/1.
Hope this helps,
and I hope you have a productive day!

8 0

3 years ago

Find the sum 2 + 4 + 6...+30

I am Lyosha [343]

Answer:

240

Step-by-step explanation:

If you are referring to:

2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30, then your answer would be 240

Hope this helps :)

6 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago

The length of a rectangular garden is 7x + 3 feet. The width of the garden is 6x – 3 feet. Which expression represents the area

iren [92.7K]

Answer:

42x² - 3x - 9

the area of the garden is:

(7x + 3)(6x - 3)

= 42x² - 21x + 18x - 9

= 42x² - 3x - 9

4 0

3 years ago

Just do part B plzz for 12 points i mark brainliest please help me!!​

Just do part B plzz for 12 points i mark brainliest please help me!!