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3 years ago

5

A customer went to a garden shop and bought some potting soil for $17.50 and 4 shrubs. The total bill was $53.50. Choose the equ

ation that models this situation and solve to find the price p of each shrub.

1 answer:

olga2289 [7]3 years ago

6 0

4x + 17.50 = 53.50

the prices of each shrub would be 9 dollars each

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3 years ago

Vector MN is comprised of the points M(8,-4) and N(-7,-2).

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3 years ago

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

On a coordinate plane, a line goes through (0, negative 3) and (3, negative 2). A point is at (negative 4, 2).

bulgar [2K]

Answer:

Correct option: first one -> y = (1/3)x + (10/3)

Step-by-step explanation:

The linear function that represents a line is:

y = ax + b

Where a is the slope and b is the y-intercept.

First we need to find the slope of the line that goes through (0, -3) and (3, -2).

Using both points, we can find the equation of the line:

x = 0 -> y = -3

-3 = a*0 + b

b = -3

x = 3 -> y = -2

-2 = 3a - 3

3a = 1

a = 1/3

The parallel line needs to have the same slope as the line, so we can model the parallel line with the following equation:

y = (1/3)x + b

The parallel line goes through the point (-4, 2), so we have:

x = -4 -> y = 2

2 = (1/3)*(-4) + b

b = 2 + (4/3)

b = 10/3

So the equation of the parallel line is:

y = (1/3)x + (10/3)

Correct option: first one

5 0

3 years ago