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grandymaker [24]
3 years ago
14

What is the range of the function f(x) = 24 + 4 when the domain is {-3, 1, 2}?

Mathematics
1 answer:
barxatty [35]3 years ago
3 0

Answer:

the 24×4 you have to count 4 times

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Gcf/ greatest common factor and / LCM least common multiple Mang Andoy is going to put in trays of 6 eggs and 12 eggs. What is t
UNO [17]

Answer with explanantion:

1. What is asked? <u>The smallest number of eggs that Mang Andoy can put using the tray.</u>

2. What are given facts?  <u>Two kinds of tray (One can have 6 eggs , other can have 12 eggs)</u>

3. How will you solve the problem? <u>We need to find LCM of 6 and 12</u>.

4. How is the solution done?

<u>Prime factorization of 6 and 12:</u>

<u>6=2 x 3</u>

<u>12 = 2 x 2 x 3</u>

<u>LCM(6,12) = 2 x 2 x 3 = 12</u>

5. What is the answer to the problem?<u> The smallest number of eggs that Mang Andoy can put using the tray = 12</u>

5 0
3 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
A company vehicle is available in black or blue.A random sample includes seven black vehicles and five blue vehicles.If there ar
hram777 [196]

Answer:

125 blue vehicles

Step-by-step explanation:

In this question, we are going to estimate the number of vehicles of a specific color given that there are some numbers of both vehicle models present in a sample.

Now to solve this question, we are going to treat the sample numbers as a ratio.

The ratio here is thus 7 black to 5 blue I.e 7:5

We have 300 vehicles, we now want to estimate the number of blue vehicles present.

From the sample ratio, there are a total of 12 vehicles

Now the total number of blue vehicles would be 5/12 * 300 = 125

There are 125 blue vehicles

8 0
3 years ago
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WILL GIVE BRAINLIEST! Is the relationship between x and y as shown in this table a proportional relationship? Explain your answe
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Answer:

No, a porportional relationship is not shown

Step-by-step explanation:

Up until 4(x) and 10(y), the relationship is porportional. Then, when x increases by 2, y also only increases by 2. Y would need to increase by 5.

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3 years ago
Mr. Singh just left his house to go to work and has a meeting as soon as he arrives. The
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Answer:5

Step-by-step explanation:

3 0
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