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dimaraw [331]
4 years ago
10

Sin a ______ = 1/2 [ sin ( a + b ) + sin ( a - b) ]

Mathematics
1 answer:
Neko [114]4 years ago
6 0

Answer:

Cos b

Step-by-step explanation:

1/2[sin(a+b)+sin(a-b)]

1/2[sin a cos b +cos a sin b + sin A cos B - cos A sin B]

1/2[2sin a cos b]

sin a cos b

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Shay makes 24 liters of homemade vegetable stock. She distributes the stock
ANTONII [103]

Answer:

3 liters

Step-by-step explanation:

pretty simple first grade question, 24/8

6 0
3 years ago
Read 2 more answers
Write 0.33 as a fraction in simplest form
Georgia [21]
33/100
this is the simplest form 
tnx
hope i hepled you

6 0
3 years ago
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What number should be placed in the box to help complete the division calculation? can someone please do the problem (8587/12) s
disa [49]

Answer:

8587 should be in the box

Step-by-step explanation:

we would see it question as:       12/- 8587

12 goes into 85  7 times, so 7

12 goes into 18 1 time, so 1

12 goes into 67 5 times, so 5

then the remainder is 7

or you further divide to get . 5 8 3333333...

so your answer should be 715r7 or 715.58333

5 0
3 years ago
$925 at 2% interest for 2.4 years. Find the simple interest earned.
mixer [17]

Answer:

£43

Step-by-step explanation:

925×2/100

=£ 18.5

2.4 years =18.5×2.4

=£43

5 0
3 years ago
A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t
Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square  at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

V(x)=x*(26-x)*(13-x)

To get maximum volume differentiate the above equation,

V(x)=x*(26*13-26*x-13*x+x^2)

V(x)=x^3-39x^2+338x\\

V'(x)=3x^2-78x+338

V''(x)=6x-78

Now ,Solve the Quadratic Equation to get x values,

3x^2-78x+338=0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17  or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0  i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

V(x)=x^3-39x^2+338x

V(x)=3.8288^3-39*(3.82883)^2+338*3.8288

V(x)=56.13-571.73+1294.13

V(x)=778.53 C.C

6 0
3 years ago
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