Which pair of angles are corresponding?

Someone find area and perimeter of this shape! Thanjs

Gemiola [76]

The area would be 132cm^2 (squared)
(9x4)x2+(7+8)x4 =132

The perimeter you just add everything up so it’s 65cm
15+13+4+7+9+4+13

6 0

3 years ago

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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

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

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

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

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

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

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

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

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

Write an equation of a line that is perpendicular to the given line and that passes through the given point.

devlian [24]

Here is the answer I believe. It goes in a x all it did was added to to each x and y (-3,0)

8 0

3 years ago

Im not understanding this question

natita [175]

It's said "The value √-9 is not -3 because --------------------

So you need to prove why -3 is not the value of √-9.

As you know (-3)^2 = (-3) * (-3) = 9, ≠ -9

Answer is A.

(-3)^2 ≠ -9

7 0

3 years ago

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(8-2•29) divided by 3

Nesterboy [21]

[(8 - 2 * 29)]/3

Let's solve inside the brackets first.

According to PEMDAS, multiplication gets solved before subtraction so multiply 2 * 29 in the parentheses.

[(8 - 58)]/3

Subtract 58 from 8 in the parentheses.

(-50)/3

Divide -50 by 3.

-16.667 or -16 2/3 is your answer.

4 0

3 years ago