Consider the functions F and G in the tables below PLEASE HELP PLEASE

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

7 0

Answer:

Option A, As x increases, the rate of change exceeds the rate of f

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Tomorrow is my exam Wow C no only

ozzi

Answer:

e. 3

f. (2x-3)(2x+3)

Step-by-step explanation:

e. 1+2=3

f. (2x)²-(3)² {a²-b²}=(a-b)(a+b)

(2x-3)(2x+3)

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3 years ago

Hey sisters i need help with this question

Allisa [31]

$3.2 \times 10^{7} + 6.4 \times 10^{7} = (3.2 + 6.4) \times 10^7 = 9.6 \times 10^7$

Answer: B

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3 years ago

Read 2 more answers

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

Lesechka [4]

Check the picture below.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{-10}~,~\stackrel{y_1}{-2})\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{4-10}{2}~~,~~\cfrac{6-2}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{4}{2} \right)\implies \stackrel{\textit{center}}{(-3~,~2)} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{\textit{center}}{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})}\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[4-(-3)]^2+[6-2]^2}\implies r=\sqrt{(4+3)^2+(6-2)^2} \\\\\\ r=\sqrt{49+16}\implies r=\sqrt{65} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{2}{ k})\qquad \qquad radius=\stackrel{\sqrt{65}}{ r} \\[2em] [x-(-3)]^2+[y-2]^2=(\sqrt{65})^2\implies (x+3)^2+(y-2)^2=65$