Select a few x x values, and plug them into the equation to find the corresponding y y values. The x x values should be selected around the vertex.
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Find the Inverse Function g(x)=5x-2. I am unable to solve this problem. g(x)=5x−2 g ( x ) = 5 x - 2. (. [. ([. ) ] )] |. |. √. √... > ≥. >≥.
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1 Answer. Jim G. Aug 14, 2018. g−1(x)=x+25. Explanation: let y=5x−2. rearrange making x the subject. 5x=y+2. x=y+25. ⇒g−1(x)=x+25.
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Inverse function of g(x)=5x−2 is g(x)=x5+2. Explanation: As g(x)=5x−2. 5x=g(x)+2 or x=15g(x)+2. Hence inverse function of g(x)=5x−2 is ...
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the question is a function ji of Axis defined as ji of X equal to minus 5 x square what is ji of -2 the question we have given that is -5 ...
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(f*g)(x) = f(x) * g(x) = 2x (5x + 2) Therefore, (f*g)(x) = 2x (5x + 2) Replace x with 3 to get the following: (f*g)(3) = 2*3* (5*3 + 2) = 6(15+2) The answer ...
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fog(x)= f(g(x))= f(|5x–2|) = ||5x–2|| =|5x–2|,Ans.
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11 thg 8, 2021 · Ex 1.3, 3Find gof and fog, if(i) f(x) = |x| and g (x) = |5x −2|f(x) = |x| , g(x) = |5x−2|f(x) = |x|f(g(x)) = |g(x)|fog(x) ...
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We want to evaluate g(-1) and then put this result back into "g". So. g(-1) = 5(-1)^2 = 5 * 1 = 5. So. g(5) = 5(5)^2 = 5*25 = 125.
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If f(x) = 2x-5x^2 and g(x) = 5x+2, what is f(4)? Possible answers include -2376, -1584, -362, -50, or none of these.
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Transcribed image text: (a) If G(X) = 5x2 – x3, find G'(a) and use it to find equations of the tangent lines to the curve y = 5x2 – x3 at the points (3, ...
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Simple and best practice solution for g(x)=5x+2 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so ...
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If f and g are two functions such that f(x) = 5x + 2 and g(x) = , then find f + g and f – g. Asked by Topperlearning User | 1st Aug, 2014, 07:51: AM.
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We will first find the value of f(g(x)), and then solve it for x = 3. If f(x) = 2x + 1 and g(x) = −5x + 2, then f(g(3)) = -25.
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We think you wrote: g(x)=5x2-4x. This solution deals with the properties of a straight line. Overview; Steps; Topics Terms and topics; Links Related links ...
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