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vfiekz [6]
3 years ago
7

Consider the functions f(x) = x2 − 13 and g(x) = x + 5. What is the value of (f − g)(−4)? ...?

Mathematics
1 answer:
MakcuM [25]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

f(x) = x2 - 13 
g(x) = x + 5

<span>and are asked to compute "f-g" when x=-4. there are two ways of doing this. you can either first find f(-4) and g(-4) and then substract one from other like so:
</span>
<span>f(−4)=(−4<span>)2</span>−13=16−13=3
</span><span>g(−4)=−4+5=1

</span>f(−4)−g(−4)=3−1=<span>2
</span>
<span>or you can create a new function that represents "f-g", call it "h", and work out h(-4) as follows:
</span>
h(x)=f(x)−g(x)=(<span>x2</span>−13)−(x+5)=<span>x2</span>−13−x−5=<span>x2</span>−x−<span>18
</span><span>h(−4)=(−4<span>)2</span>−(−4)−18=16+4−18=2</span>
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Read 2 more answers
Show work and explain with formulas:
Marina86 [1]

17 Answer:  205

<u>Step-by-step explanation:</u>

\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}

18 Answer:  1968

<u>Step-by-step explanation:</u>

a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}

19 Answer:  -116

<u>Step-by-step explanation:</u>

\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}

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