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3 years ago

What is the area of a rectangle that is 12 cm long and 2 cm wide?

Mathematics

2 answers:

Hoochie [10]3 years ago

4 0

Area = Width x Height
12 x 2 = 24
24 cm is your answer

sdas [7]3 years ago

3 0

24 cm is your answer. It's 12*2.

You might be interested in

Which table represents a function? A 2-column table with 4 rows. The first column is labeled x with entries negative 3, 0, negat

tangare [24]

Answer:

This is a many-to-one function as the value of y = -1 corresponds to two values of x:

$\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -3 & -1\\\cline{1-2} 0 & 0\\\cline{1-2} -2 & -1\\\cline{1-2} 8 & 1\\\cline{1-2}\end{array}$

Step-by-step explanation:

Function

A special relationship where each input (x-value) has a single output (y-value).

A function is one-to-one if each value in the range (y-values) corresponds to exactly one value in the domain (x-values).

A function is many-to-one if some values in the range (y-values) correspond to more than one (many) value in the domain (x-values).

This is a many-to-one function as the value of y = -1 corresponds to two values of x:

$\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -3 & -1\\\cline{1-2} 0 & 0\\\cline{1-2} -2 & -1\\\cline{1-2} 8 & 1\\\cline{1-2}\end{array}$

This is not a function as the value of x = -5 corresponds to two values of y:

$\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -5 & -5\\\cline{1-2} 0 & 0\\\cline{1-2} -5 & 5\\\cline{1-2} 6 & -6\\\cline{1-2}\end{array}$

This is not a function as the value of x = -2 corresponds to two values of y:

$\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -4 & 8\\\cline{1-2} -2 & 2\\\cline{1-2} -2 & 4\\\cline{1-2} 0 & 2\\\cline{1-2}\end{array}$

This is not a function as the value of x = -4 corresponds to two values of y:

$\begin{array}{|c|c|}\cline{1-2} x & y\\\cline{1-2} -4 & 2\\\cline{1-2} 3 & 5\\\cline{1-2} 1 & 3\\\cline{1-2} -4 & 0\\\cline{1-2}\end{array}$

8 0

2 years ago

Question 1

drek231 [11]

QUESTION 1

We want to expand $(x-2)^6$ .

We apply the binomial theorem which is given by the formula

$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+...+^nC_na^{n-n}b^n$ .

By comparison,

$a=x,b=-2,n=6$ .

We substitute all these values to obtain,

$(x-2)^6=^6C_0x^6(-2)^0+^6C_1x^{6-1}(-2)^1+^6C_2x^{6-2}(-2)^2+^6C_3x^{6-3}(-2)^3+^6C_4x^{6-4}(-2)^4+^6C_5x^{6-5}(-2)^5+^6C_6x^{6-6}(-2)^6$ .

We now simplify to obtain,

$(x-2)^6=^nC_0x^6(-2)^0+^6C_1x^{5}(-2)^1+^6C_2x^{4}(-2)^2+^6C_3x^{3}(-2)^3+^6C_4x^{2}(-2)^4+^6C_5x^{1}(-2)^5+^6C_6x^{0}(-2)^6$ .

This gives,

$(x-2)^6=x^6-12x^{5}+60x^{4}-160x^{3}(-2)^3+240x^{2}-1925x+64$ .

Ans:C

QUESTION 2

We want to expand

$(x+2y)^4$ .

We apply the binomial theorem to obtain,

$(x+2y)^4=^4C_0x^4(2y)^0+^4C_1x^{4-1}(2y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(2y)^3+^4C_4x^{4-4}(2y)^4$ .

We simplify to get,

$(x+2y)^4=x^4(2y)^0+4x^{3}(2y)^1+6x^{2}(2y)^2+4x^{1}(2y)^3+x^{0}(2y)^4$ .

We simplify further to obtain,

$(x+2y)^4=x^4+8x^{3}y+24x^{2}y^2+32x^{1}y^3+16y^4$

Ans:B

QUESTION 3

We want to find the number of terms in the binomial expansion,

$(a+b)^{20}$ .

In the above expression, $n=20$ .

The number of terms in a binomial expression is $(n+1)=20+1=21$ .

Therefore there are 21 terms in the binomial expansion.

Ans:C

QUESTION 4

We want to expand

$(x-y)^4$ .

We apply the binomial theorem to obtain,

$(x-y)^4=^4C_0x^4(-y)^0+^4C_1x^{4-1}(-y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(-y)^3+^4C_4x^{4-4}(-y)^4$ .

We simplify to get,

$(x+2y)^4=^x^4(-y)^0+4x^{3}(-y)^1+6x^{2}(-y)^2+4x^{1}(-y)^3+x^{0}(-y)^4$ .

We simplify further to obtain,

$(x+2y)^4=x^4-4x^{3}y+6x^{2}y^2-4x^{1}y^3+y^4$

Ans: C

QUESTION 5

We want to expand $(5a+b)^5$

We apply the binomial theorem to obtain,

$(5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{5-1}(b)^1+^5C_2(5a)^{5-2}(b)^2+^5C_3(5a)^{5-3}(b)^3+^5C_4(5a)^{5-4}(b)^4+^5C_5(5a)^{5-5}(b)^5$ .

We simplify to obtain,

$(5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{4}(b)^1+^5C_2(5a)^{3}(b)^2+^5C_3(5a)^{2}(b)^3+^5C_4(5a)^{1}(b)^4+^5C_5(5a)^{0}(b)^5$ .

This finally gives us,

$(5a+b)^5=3125a^5+3125a^{4}b+1250a^{3}b^2+^250a^{2}(b)^3+25a(b)^4+b^5$ .

Ans:B

QUESTION 6

We want to expand $(x+2y)^5$ .

We apply the binomial theorem to obtain,

$(x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{5-1}(2y)^1+^5C_2(x)^{5-2}(2y)^2+^5C_3(x)^{5-3}(2y)^3+^5C_4(x)^{5-4}(2y)^4+^5C_5(x)^{5-5}(2y)^5$ .

We simplify to get,

$(x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{4}(2y)^1+^5C_2(x)^{3}(2y)^2+^5C_3(x)^{2}(2y)^3+^5C_4(x)^{1}(2y)^4+^5C_5(x)^{0}(2y)^5$ .

This will give us,

$(x+2y)^5=x^5+^10(x)^{4}y+40(x)^{3}y^2+80(x)^{2}y^3+80(x)y^4+32y^5$ .

Ans:A

QUESTION 7

We want to find the 6th term of $(a-y)^7$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=7,b=-y$

We substitute to obtain,

$T_{5+1}=^7C_5a^{7-5}(-y)^5$ .

$T_{6}=-21a^{2}y^5$ .

Ans:D

QUESTION 8.

We want to find the 6th term of $(2x-3y)^{11}$

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=11,a=2x,b=-3y$

We substitute to obtain,

$T_{5+1}=^{11}C_5(2x)^{11-5}(-3y)^5$ .

$T_{6}=-7,185,024x^{6}y^5$ .

Ans:D

QUESTION 9

We want to find the 6th term of $(x+y)^8$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=8,a=x,b=y$

We substitute to obtain,

$T_{5+1}=^8C_5(x)^{8-5}(y)^5$ .

$T_{6}=56a^{3}y^5$ .

Ans: A

We want to find the 7th term of $(x+4)^8$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=6,n=8,a=x,b=4$

We substitute to obtain,

$T_{6+1}=^8C_5(x)^{8-6}(4)^6$ .

$T_{7}=114688x^{2}$ .

Ans:A

4 0

3 years ago

What is the inverse of the function f(x) = x +3?

KatRina [158]

Next time please add answer choices

f(x) = y = x+3, Consider y=x+3. You just switch x and y:
so y= x+3, becomes x=y+3. Once done, you isolate y:
x=y+3 or y = x-3. This this the inverse function of f(x)

5 0

3 years ago

sarah used 3/4 pound of blueberries to make 1/2 cup of jam. how many pounds of blueberries would she need to make a cup of jam?

Ksenya-84 [330]

Answer: She needed $1\frac{1}{2}\text{ pounds of blueberries}$ to make a cup of jam .

Step-by-step explanation:

Since we have given that

Sarah used $\frac{3}{4}$ pound of blueberries to make $\frac{1}{2}$ cup of jam.

As we need to find the number of pounds of blueberries she would need to make a cup of jam.

So, we will use "Unitary Method":

$\text{For }\frac{1}{2}\text{ cup of jam},\\\\\text{ she need }\frac{3}{4}\text{ pound of blueberries }\\\\\text{ For 1 cup of jam}.\\\\\text{She need }=\frac{3}{4}\times 2=\frac{3}{2}=1\frac{1}{2}\text{ pounds of blueberries }$

Hence, She needed $1\frac{1}{2}\text{ pounds of blueberries}$ to make a cup of jam .

4 0

3 years ago

Solve –2x2 +3x – 9 = 0. (solving quadratic equation with complex numbers)..

goldfiish [28.3K]

-3+-sqrt9-56/-4
-3+-sqrt-47/-4
-3+-sqrt47i/-4

4 0

3 years ago