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stepan [7]
3 years ago
9

8g+7+3g+12 can be written as?​

Mathematics
2 answers:
umka21 [38]3 years ago
8 0

Answer:

It can be also written as  11g+19.

Step-by-step explanation:

First, you would add the numbers with the 'g'

11g+7+12

then you would add the regular numbers.

11g+19

jenyasd209 [6]3 years ago
4 0

Answer:

11g + 19

Hope this helps

Step-by-step explanation:

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Please, I need help in this ??
nignag [31]

Answer:

\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c

Step-by-step explanation:

\int\frac{x^{4}}{x^{4} -1}dx

Adding and Subtracting 1 to the Numerator

\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx

Dividing Numerator seperately by x^{4} - 1

\int 1 + \frac{1}{x^{4}-1 }\, dx

Here integral of 1 is x +c1 (where c1 is constant of integration

x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx----------------------------------(1)

We apply method of partial fractions to perform the integral

\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}------------------------------------------(2)

\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}

1 = A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)-------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = \frac{1}{4}

1 = B(-1-1)(1+1)

B = -\frac{1}{4}

1 = C(i-1)(i+1)

C = -\frac{1}{2}

Substituting A, B, C in equation (2)

\int\frac{x^{4}}{x^{4} -1}dx = \int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }

On integration

Here \int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx

\int\frac{x^{4}}{x^{4} -1}dx = \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2

Here c1 + c2 can be added to another and written as c

Therefore,

\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c

4 0
3 years ago
Write the 5th term in the expansion of (5x-y/2)^7
Nezavi [6.7K]

Answer:

65625/4(x^5)(y²)

Step-by-step explanation:

Using binomial expansion

Formula: (n k) (a^k)(b ^(n-k))

Where (n k) represents n combination of k (nCk)

From the question k = 5 (i.e. 5th term)

n = 7 (power of expression)

a = 5x

b = -y/2

....................

Solving nCk

n = 7

k = 5

nCk = 7C5

= 7!/(5!2!) ------ Expand Expression

=7 * 6 * 5! /(5! * 2*1)

= 7*6/2

= 21 ------

.........................

Solving (a^k) (b^(n-k))

a = 5x

b = -y/2

k = 5

n = 7

Substituting these values in the expression

(5x)^5 * (-y/2)^(7-5)

= (3125x^5) * (-y/2)²

= 3125x^5 * y²/4

= (3125x^5)(y²)/4

------------------------------------

Multiplying the two expression above

21 * (3125x^5)(y²)/4

= 65625/4(x^5)(y²)

5 0
3 years ago
Find the value of x in the triangle shown below
olga nikolaevna [1]

Answer:

102

Step-by-step explanation:

it is an isosceles triangle as two sides are equal i.e 3.3 so angles opposite to equal sides are also equal so 39 plus 39 plus x = 180°

78 plus x =180

x=180-78

x=102°

hope it helps

8 0
3 years ago
Sorry here's another problem I need help with​
tiny-mole [99]

Answer:

32

Step-by-step explanation:

You need to work backwards to solve this equation!

9 - 17 = -8

w / -4 = -8

32 / -4 = -8

Which means that  w = 32

5 0
3 years ago
I don't understand and I need this done someone please help
Hatshy [7]
Divide them and then multiply
5 0
3 years ago
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