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Alexeev081 [22]

3 years ago

7

A builder has agreed not to erect all "look alike" homes in a new subdivision. Five exterior designs are offered to potential ho

mebuyers. The builder has standardized three interior plans that can be incorporated in any of the five exteriors. How many different ways can the exterior and interior plans be offered to potential homebuyers?

1 answer:

shutvik [7]3 years ago

4 0

Answer:

There will be 15 ways.

Step-by-step explanation:

5 exterior designs are offered to potential home buyers.

The builder has standardized 3 interior plans that can be incorporated in any of the 5 exteriors.

So, the number of different ways in which the exterior and interior plans be offered to potential home buyers is given by:

$5\times3=15$

Therefore, the answer is 15 ways.

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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

Please help me 50 points

Afina-wow [57]

angle 3 = angle 6 = 60°

hope it helps...!!!

3 0

2 years ago

Read 2 more answers

Write a linear equation with a slope of 14 and y-intercept -12.

Vanyuwa [196]

Answer:

y=14x-12

Step-by-step explanation:

in the standard format y=mx+b, m is the slope and b is the y-intercept, so this equation would be y=14x-12 since the slope is 14 and the y-intercept is -12.

8 0

3 years ago

mr Goodwill [35]

Answer:

34 Braids

Step-by-step explanation:

.85 per braid means 1 is 85 cents, 2 is $1.70 and so on. So normally you would just divide the total amount of money, so for instance if you had $2.55 you would be able to afford 3 braids, since 3 braids cost $2.55. in fact $2.56, $2.57 aaaaaaaaaaall the way up to $3.39 would be 3 braids, then $3.40 would be 4 braids.

Here though you first pay 3 dollars. so you need 3 first, then an extra 85 cents to get 1 braid. 2 braids is 3 dollars plus 1.70, 3 braids is 3 + 2.55 4 braids is 3 + 3.40 and so on.

So what you want to do is first take away that 3 dollars you need before the braids, then divide by .85. for instance $6.40 would get you 4 braids as I showed bcause 3 + 3.40 = 6.40. now 6.40 - 3 = 3.40 then 3.40 divided by .85 = 4

The problem says we start with$32, so lets do those steps. first 32 - 3 = 29. then 29 divided by .85 = 34.11. well it's not whole number so lets try something. 34 braids would be 3 + .85*34 = 31.90 and 35 braids would be 32.75 So 31.90, 31.91 aaaaaaaaall the way up to 32.74 would be 34 braids then 32.75 would be 35 braids. well, $32 is in that in between space, so it is 34 braids.

When I did 32 - 3 = 29. then 29 divided by .85 = 34.11 that 34.11 does actually tell us we wouldn't get to 35, I just wanted to make it very clear. so the answer is 34 braids.

7 0

2 years ago

Mr. Daniel has 1 2/3 acres of land in which he plans to fertilize. If each bag of fertilizer covers 1/9 of an acre and be bought

tensa zangetsu [6.8K]

He will have 5 bags leftover.

7 0

2 years ago

Read 2 more answers