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3 years ago

List all the 4 digit numbers that fit these clues

Mathematics

1 answer:

natita [175]3 years ago

4 0

If you give the clues I could help.

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6x - 5y = 15, x=y+3 what does x and y equal

Natasha_Volkova [10]

Answer:

Step-by-step explanation:

6x - 5y = 15....... (1)

x=y+3..........(2)

Putting (2) in (1)

6(y + 3) - 5y = 15

6y + 18 - 5y = 15

6y - 5y = 15 - 18

y = -3

And x = y + 3

x = -3 + 3

x = 0

3 0

3 years ago

Plzz help will mark brainiest Don't do it for the points

UNO [17]

Answer:

I think the answer is B

Step-by-step explanation:

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3 years ago

What is the scale factor from figure A to figure B?

SCORPION-xisa [38]

Answer:

Figure A sides are 1/4 the size of Figure B

The top of Figure A = 4, the top of Figure B = 1.

Divide 1 by 4 to get the scale factor, which is 1/4 as a fraction or 0.25 as a decimal.

Step-by-step explanation:

3 0

3 years ago

You roll a fair 6-sided die.

Yuki888 [10]

Answer:

1/3

Step-by-step explanation:

7 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

3 years ago