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2 years ago

For f(x) = 4x+1 and g(x)=x²-5, find (f-g)(x).

Mathematics

2 answers:

Katen [24]2 years ago

5 0

Answer:

(f-g) (x) = 4x+1 - x² + 5 = -x² + 4x + 6

Step-by-step explanation:

Explanation can be given if needed in the comments (ㅅ´ ˘ `)

SSSSS [86.1K]2 years ago

4 0

Answer: -x² + 4x + 6

Given:

f(x) = 4x + 1

g(x) = x² - 5

Solve:

= (f - g)(x)

= f(x) - g(x)

= 4x + 1 - (x² - 5)

= 4x + 1 - x² + 5

= -x² + 4x + 6

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4 years ago

PLEASE HELP ASAP -- WITH THE WORKK<br><br>50 points

MAXImum [283]

Part-A:-

$\\ \sf\longmapsto x^2+3x-70(4x-12)$

$\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840$

$\\ \sf\longmapsto -8x^2+48x+560$

Now

$\\ \sf\longmapsto (f.g)(3)$

$\\ \sf\longmapsto -8(3)^2+48(3)+560$

$\\ \sf\longmapsto -72+144+560$

$\\ \sf\longmapsto 72+560$

$\\ \sf\longmapsto 632$

Part-B:-

$\\ \sf\longmapsto f(3)$

$\\ \sf\longmapsto (3)^2+3(3)-70$

$\\ \sf\longmapsto 9+9-70$

$\\ \sf\longmapsto 18-70$

$\\ \sf\longmapsto -52$

Now

$\\ \sf\longmapsto g(f(3))$

$\\ \sf\longmapsto g(-52)$

$\\ \sf\longmapsto 4(-52-3)$

$\\ \sf\longmapsto 4(-55)$

$\\ \sf\longmapsto -220$

Part:-C

$\\ \sf\longmapsto g(-2)$

$\\ \sf\longmapsto 4(-2-3)$

$\\ \sf\longmapsto 4(-5)$

$\\ \sf\longmapsto -20$

Now

$\\ \sf\longmapsto f(g(-2))$

$\\ \sf\longmapsto f(-20)$

$\\ \sf\longmapsto (-20)^2+3(-20)-70$

$\\ \sf\longmapsto 400-60-70$

$\\ \sf\longmapsto 400-130$

$\\ \sf\longmapsto 270$

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g×f=f×g

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$\\ \sf\longmapsto -8(-2)^2+48(-2)+560$

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3 years ago

18. Determine the common difference, the fifth term, and the sum of the first 100 terms of the following sequence:

Ksenya-84 [330]

$a_1=1,\ a_2=2.5,\ a_3=4,\ a_4=5.5,\ ...\\\\a_2-a_1=2.5-1=1.5\\a_3-a_2=4-2.5=1.5\\a_4-a_3=5.5-4=1.5\\a_{n+1}-a_n=1.5=constans\\\\\text{It's an arithmetic sequence with}\\a_1=1,\ \boxed{d=1.5}\\\\a_n=a_1+(n-1)d\to a_n=1+(n-1)(1.5)=1+1.5n-1.5\\\\a_n=1.5n-0.5\\\\a_5=1.5(5)-0.5=7.5-0.5=7\\\boxed{a_5=7}\\\\\text{The formula of a Sum of the First n Terms of an Arithmetric Sequence:}\\\\S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n\\\\\text{We have:}\\a_1=1,\ d=1.5,\ n=100\\\\\text{Substitute}$

$S_{100}=\dfrac{(2)(1)+(100-1)(1.5)}{2}\cdot100=\dfrac{2+(99)(1.5)}{1}\cdot50\\\\=(2+148.5)\cdot50=150.5\cdot50=7,525\\\\\boxed{S_{100}=7,525}\\\\Answer:\\the\ common\ difference:\ d=1.5\\the\ fifth\ term:\ a_5=7\\the\ sum\ of\ first\ 100\ terms:\ S_{100}=7,525$

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3 years ago