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erastovalidia [21]
3 years ago
7

does anyone know a good way to remember the difference between the domain and range?? if so can you tell me??

Mathematics
2 answers:
lara [203]3 years ago
7 0
The domain are the x values and the range is the opposite of that which are the y values
Basile [38]3 years ago
6 0
The domain is the values that x is allowed to be and range is the values that the y's end up being
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What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

<u>Solution: </u>

The given expression is (2 x-3 y)^{10}

As per binomial theorem, we know,

(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now x^{5}\times y^5 will be the 6 term where k =5

Now, \mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}

So, the coefficient of x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right.

10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252

The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

6 0
3 years ago
Multiply. Express your answer as a fraction in<br> Simplest form<br> 8.33 x V 144
adell [148]

Answer

65-44.78>4

Step-by-step explanation

4 1
3 years ago
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Find the difference 9x2-24x2
vovikov84 [41]
So, I'm assuming that the x represents multiplication...
Because multiplication has a higher precedence over addition, we'll multiply the numbers first:
18 + 48
Then we simply add:
66
6 0
3 years ago
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Complete the statement of each of these rules:
never [62]

Answer:

a) (+) × (?) = (+)

(?) = (+)

b) (-) × (+) = ?

? = (-)

c) (?) × (-) = ?

? is undefined

Step-by-step explanation:

a) (+) × (?) = (+)

From the given equation in the question, by dividing both side of the equation by (+), we have;

((+) × (?))/(+) = (+)/(+)

(?) = (+)/(+) = (+)

Therefore, we have;

(+) × (+) = (+)

(?) = (+)

b) (-) × (+) = ?

From the given equation in the question we apply the operation rules as follows;

(-) × (-) = (+)

(-) × (+) = (-)

Therefore;

(-) × (+) = ? = (-)

? = (-)

c) (?) × (-) = ?

From the rules of multiplication, we have;

(-) × (-) = (+)

(+) × (-) = (-)

Therefore, when (?) = (-), ? = +, from which we have;

(?) ≠ ?, which is an error

Similarly, when (?) = (+), ? = -, therefore, (?) and ?, always changes sign

Therefore, the equation, (?) × (-) = ?, is undefined.

3 0
3 years ago
Brian is solving the equation x^2 - 3/4x= 5. What value must be added to both sides of the equation to make the left side a perf
Yakvenalex [24]
<span>((-3/4)/2)^2 = 9/64 I HOPE THIS HELP</span>
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