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1 year ago

8

How many positive odd integers less than 10,000 can be written using the digits

1 answer:

Ostrovityanka [42]1 year ago

5 0

125 positive odd integers

Explanation:

I was given this question before and remember it. I assure you 125 is the correct answer.

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What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

When planning a date night, you have a choice of 2 types of restaurants: pizza (P) or barbeque (B); a choice of 3 types of movie

enot [183]

Answer:

$S = \{PRC, PRI,PAC,PAI,PDC,PDI,BRC, BRI,BAC,BAI,BDC,BDI\}$

Step-by-step explanation:

Given

$Restaurant = \{P,B\}$

$Movies = \{R,A,D\}$

$Post-movie = \{C,I\}$

Required

List out the sample space

To do this, we simply pick 1 item in each category, without repetition.

So, we have:

$S = \{PRC, PRI,PAC,PAI,PDC,PDI,BRC, BRI,BAC,BAI,BDC,BDI\}$

4 0

2 years ago

Given points A(-1, -2) and B(2, 4) where AP: BP=1:2, find the locus of point P.

koban [17]

Answer:

$x^2 + 4x + y^2 +8y = 0$

Step-by-step explanation:

Given

$A = (-1,-2)$

$B = (2,4)$

$AP:BP = 1 : 2$

Required

The locus of P

$AP:BP = 1 : 2$

Express as fraction

$\frac{AP}{BP} = \frac{1}{2}$

Cross multiply

$2AP = BP$

Calculate AP and BP using the following distance formula:

$d = \sqrt{(x - x_1)^2 + (y - y_1)^2}$

So, we have:

$2 * \sqrt{(x - -1)^2 + (y - -2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

$2 * \sqrt{(x +1)^2 + (y +2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

Take square of both sides

$4 * [(x +1)^2 + (y +2)^2] = (x - 2)^2 + (y - 4)^2$

Evaluate all squares

$4 * [x^2 + 2x + 1 + y^2 +4y + 4] = x^2 - 4x + 4 + y^2 - 8y + 16$

Collect and evaluate like terms

$4 * [x^2 + 2x + y^2 +4y + 5] = x^2 - 4x + y^2 - 8y + 20$

Open brackets

$4x^2 + 8x + 4y^2 +16y + 20 = x^2 - 4x + y^2 - 8y + 20$

Collect like terms

$4x^2 - x^2 + 8x + 4x + 4y^2 -y^2 +16y + 8y + 20 - 20 = 0$

$3x^2 + 12x + 3y^2 +24y = 0$

Divide through by 3

$x^2 + 4x + y^2 +8y = 0$

3 0

3 years ago

Zorna weighs 92 pounds. Her weight is 6 pounds more than half of her father's weight. Which equation would help me find out how

8_murik_8 [283]

$92 - 6 = 86$

4 0

3 years ago

Please help due in 10 minutes

Anit [1.1K]

3(4x - 4) - 7x, when simplified, is 5x - 12. So your answer would be A, or the first one.

4 0

2 years ago

Read 2 more answers