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natita [175]
3 years ago
5

HELP I WILL MAEK BEAINLIST PLEASE

Mathematics
1 answer:
wolverine [178]3 years ago
6 0

Answer:

260,548668 or 260.55 round to the nearest hundreds  

Step-by-step explanation:

Area of the trapezoid:

[(major base + minor base)x height]/2

[(22 + 12) x 12]/2 = 204 cm^2

radius = diameter / 2 = 12 / 2 = 6 cm

Area of a semicircle

(radius^2 x pi)/2 = (6^2 x pi)/2 = 56.548668 cm^2

total area = 204 + 56,548668 = 260,548668 cm^2

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H(x)=-x^2+10x-16 Find the 2 x-intercepts
Katena32 [7]

Answer:

x=2   x=8

Step-by-step explanation:

h(x)=-x^2+10x-16

Set this equal to 0

0=-x^2+10x-16

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0/-1=x^2-10x+16

0=x^2-10x+16

Factor

What two numbers multiply to 16 and add to -10

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Using the zero product property

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7 0
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Pond A and Pond B have 20 fishes altogether.
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Answer:

29

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3 0
3 years ago
Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B​
telo118 [61]

In order to prove

\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)

Let's write both sides in terms of \sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x) only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

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

and

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

So, their multiplication is

\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)

So, the left hand side simplifies to

\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now, on with the right hand side. We have

1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now simply make this expression one fraction:

1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

And as you can see, the two sides are equal.

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600 over 1000 converted to simplest from
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0.6 or 3/5
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