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3 years ago

9

Need help on this two column proof for geometry

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

e=mc

Step-by-step explanation:

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Finley’s pumpkin had a mass of 6.5 kilograms before he carved it. After carving it the pumpkin had a mass of 3.9 kilograms. What

katen-ka-za [31]

Hello.

The percent decrease in the mass of the pumpkin for this case will be given by:
(100 - (3.9 / 6.5) * 100) =
(100-60) =40%

After carving it, the pumpkin decreases a percent of 40% in the mass.

Have a nice day

4 0

2 years ago

Read 2 more answers

What did the ancestral puebloans do to make the growing need for housing

koban [17]

Answer: How did the Ancestral Puebloans adapt their housing to the building materials that were available in the arid Southwest? They built some houses of stone blocks against canyon walls and built freestanding villages of stone and adobe brick.

Step-by-step explanation:

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3 years ago

Arrange from least to greatest.<br> 5/6 2/4 5/12

Jet001 [13]

Answer:

5/6=10/12. 2/4=6/12. 5/12

5/12. 2/4. 5/6

6 0

2 years ago

What is 13x + 21y - 14x - y simplified?

Arisa [49]

Answer:

-x + 20y

Step-by-step explanation:

trust me.. its right

4 0

2 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

2 years ago