All categories

2 years ago

Will give brainliest please help ASAP

Mathematics

1 answer:

jolli1 [7]2 years ago

8 0

The inequality sign is greater than or equal to, which means the line would be solid and the shaded portion would be to the right of the line.

The correct graph would be A.

You might be interested in

PLEASE HELP!! I’ll mark brainliest!! Image of the figure will be attached. 2.5 2.5 and 5 are the numbers! To get brainliest plea

balandron [24]

Answer:

109.375 cm³

Step-by-step explanation:

The volume of the cube is:

V = a³ = 5³ = 125 cm³

The volume of the cut prism:

V = Bh, B- area of the base, h- height

The base is a right triangle with legs of 2.5 each.

The volume is:

V = 1/2*2.5*2.5*5 = 15.625 cm³

Required volume of the solid is the difference of the above volumes:

125 - 15.625 = 109.375 cm³

5 0

3 years ago

Read 2 more answers

Larry earns $11.25/hour selling televisions. He also earns 4.5% commission on his sales. What is his

monitta

Answer:

C) $573.75

Step-by-step explanation:

We can come up with the equation of this Gross Weekly pay as:

$P(t)=11.25t+0.045S$

Where P is the amount Larry will get paid, t is the time worked and S is the amount of sales he generated that week. This was derived from the fact that 11.25 is the slope and 0.045S is the y-intercept. By plugging in those values we have:

$P(35)=11.25(35)+0.045(4000)=573.75$

C) $573.75

4 0

2 years ago

Will you help me on both

NeTakaya

Answer:R

Step-by-step explanation:

6 0

2 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

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

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

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

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

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

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

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