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3 years ago

15

Function 1: y = 4x + 5

1 answer:

evablogger [386]3 years ago

6 0

Answer: function 1

Rate of change of function 1:
Following the format of y=mx+c, the rate of change should be m, so, the rate of change for function 1 = 4

To find the gradient (rate of change):
The two points the line passes through are (x1, y1) and (x2, y2), which in this case is (1, 6) and (3, 10)
(Doesn't matter which is which but you need to make sure that once you decide which is which, you stick to it)
To calculate the gradient, you substitute these values following (y1 - y2)/(x1 - x2)

Gradient of function 2 = (10 - 6)/(3 - 1)
= 2

Therefore, since 4 > 2, rate of change of function 1 > rate of change of function 2.

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If 9:3 is equal to 3:1 and then what is 12:4 equal to ?

avanturin [10]

Answer:

36:12 or 3:1

Step-by-step explanation:

12:4= 36: 12

or 12:4 simplify= 3:1

hope it helps

4 0

2 years ago

1+9x14+12=? please help fast

Klio2033 [76]

Answer:

139

Step-by-step explanation:

So we start with PEMDAS. (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.)

Since we don't have parenthesis or exponents, we can go straight to multiplication. 9 multiplied by 14 is 126.

Now we go to addition. 126+1=127

127+12=139

Hope I helped :)

~Sunflower UwU

5 0

2 years ago

Read 2 more answers

If f(x)=x^2-5x+4, if is possible to find f(x+4) by replacing each occurrence of

shusha [124]

Answer

It is possible to find f(x+7) by replacing each occurrence of

X by X+7

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2 years ago

Find simple interest amount for 1000 for one year at 6%

DIA [1.3K]

Answer:

The ans is in the picture with the steps how i got it

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Step-by-step explanation:

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2 years ago

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

2 years ago