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ASHA 777 [7]
2 years ago
6

Is 3n a cubic a quadratic a linear or none of these​

Mathematics
1 answer:
lilavasa [31]2 years ago
7 0

Step-by-step explanation:

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Class 10

>>Maths

>>Polynomials

>>Revisiting Polynomials

>>Classify the following as linear, quadra

Question

Bookmark

Classify the following as linear, quadratic and cubic polynomials:

(i) x

2

+x

(ii) x−x

3

(Iii) y+y

2

+4

(iv) 1+x

(v) 3t

(vi) r

2

(vii) 7x

3

Medium

Solution

verified

Verified by Toppr

(i) The highest degree of x

2

+x is 2, so it is a quadratic polynomials.

(ii) The highest degree of x−x

3

is 3, so it is a cubic polynomials.

(iii)The highest degree of y+y

2

+4 is 2, so it is a quadratic polynomials.

(iv) The highest degree of x in (1+x) is 1, so it is a linear polynomials.

(v)The highest degree of t in 3t is 1, so it is a linear polynomials.

(vi)The highest degree of r

2

is 2, so it is a quadratic polynomials.

(vii)The highest degree of x in 7x

3

is 3, so it is a cubic polynomials.

Video Explanation

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4 0
3 years ago
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A new shopping mall records 120 total shoppers on their first day of business. Each day after that, the number of shoppers is 10
Iteru [2.4K]

Answer:

1,139\ shoppers

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

y=a(b)^{x}

where

a is the initial value or y-intercept

b is the base of the exponential function

r is the rate in decimal form

b=(1+r)

In this problem we have

x ----> the number of days

y ----> the number of shoppers

a=120

r=10%=10/100=0.10

b=1+0.10=1.10

substitute the values

y=120(1.10)^{x}

First day

y=120

Second day

For x=1 day

substitute the value of x in the equation and solve for y

y=120(1.10)^{1}=132

Third day

For x=2 days

y=120(1.10)^{2}=145

Fourth day

For x=3 days

y=120(1.10)^{3}=160

Fifth day

For x=4 days

y=120(1.10)^{4}=176

Sixth day

For x=5 days

y=120(1.10)^{5}=193

Seventh day

For x=6 days

y=120(1.10)^{6}=213

Adds the numbers

120+132+145+160+176+193+213=1,139

3 0
3 years ago
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