https://matematikk.net/w/api.php?action=feedcontributions&feedformat=atom&user=ThomasSkas 2026-04-04T10:30:07Z Brukerbidrag MediaWiki 1.42.3 https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12564 2014-02-23T17:19:21Z

<p>ThomasSkas: /* Oppgave 8 */</p> <hr /> <div>==DEL EN==<br /> <br /> ==Oppgave 1==<br /> <br /> $f(x)= 3x^2+3x+1 \qquad D_f = \R \\f'(x)= 6x-3 \\f´(2)= 6 \cdot 2 -3 = 9$<br /> <br /> ==Oppgave 2==<br /> <br /> <br /> ===a)===<br /> $x(x+5)-10 =4 \\ x^2+5x-14=0 \\ x= \frac{-5\pm \sqrt{25+56}}{2} \\x= -7 \vee x=2$<br /> <br /> ===b)===<br /> <br /> $10^{3x} -100000 =0 \\ 10^{3x} = 10^5 \\ 3x=5 \\ x=\frac 53$<br /> <br /> ==Oppgave 3==<br /> ==Oppgave 4==<br /> <br /> <br /> ===a)===<br /> $v=v_0+at \\t = \frac{v-v_0}{a}$<br /> <br /> ===b)===<br /> $t = \frac{v-v_0}{a} \\<br /> t = \frac{25-1}{3}=8 $<br /> <br /> ==Oppgave 5==<br /> <br /> ===a)===<br /> <br /> $ \frac{9^2a^2b^3}{(3ab^2)^2} \\ =\frac{3^4a^2b^3}{3^2a^2b^6} \\ = 3^{4-2} \cdot a^{2-2} \cdot b^{3-6} \\ = 3^2b^{-3} \\ = \frac{9}{b^3} $<br /> <br /> ===b)===<br /> <br /> $\lg (\frac{a^2}{b^2}) + \lg( \frac {b^2}{a}) + lg (a+b) \\ =\lg a^2 - \lg b^2 + \lg b^2 - \lg a + \lg (a+b) \\= 2\lg a - \lg a + \lg(a+b) \\ = \lg a + \lg (a+b) \\ =lg (a(a+b)) \\ =\lg (a^2+ab)$<br /> <br /> ==Oppgave 6==<br /> ==Oppgave 7==<br /> ==Oppgave 8==<br /> \sqrt{a^2+b^2}<br /> <br /> ==Oppgave 9==<br /> <br /> ==Oppgave 10==</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12563 2014-02-22T21:06:32Z

<p>ThomasSkas: /* Oppgave 8 */</p> <hr /> <div>==DEL EN==<br /> <br /> ==Oppgave 1==<br /> <br /> $f(x)= 3x^2+3x+1 \qquad D_f = \R \\f'(x)= 6x-3 \\f´(2)= 6 \cdot 2 -3 = 9$<br /> <br /> ==Oppgave 2==<br /> <br /> <br /> ===a)===<br /> $x(x+5)-10 =4 \\ x^2+5x-14=0 \\ x= \frac{-5\pm \sqrt{25+56}}{2} \\x= -7 \vee x=2$<br /> <br /> ===b)===<br /> <br /> $10^{3x} -100000 =0 \\ 10^{3x} = 10^5 \\ 3x=5 \\ x=\frac 53$<br /> <br /> ==Oppgave 3==<br /> ==Oppgave 4==<br /> <br /> <br /> ===a)===<br /> $v=v_0+at \\t = \frac{v-v_0}{a}$<br /> <br /> ===b)===<br /> $t = \frac{v-v_0}{a} \\<br /> t = \frac{25-1}{3}=8 $<br /> <br /> ==Oppgave 5==<br /> <br /> ===a)===<br /> <br /> $ \frac{9^2a^2b^3}{(3ab^2)^2} \\ =\frac{3^4a^2b^3}{3^2a^2b^6} \\ = 3^{4-2} \cdot a^{2-2} \cdot b^{3-6} \\ = 3^2b^{-3} \\ = \frac{9}{b^3} $<br /> <br /> ===b)===<br /> <br /> $\lg (\frac{a^2}{b^2}) + \lg( \frac {b^2}{a}) + lg (a+b) \\ =\lg a^2 - \lg b^2 + \lg b^2 - \lg a + \lg (a+b) \\= 2\lg a - \lg a + \lg(a+b) \\ = \lg a + \lg (a+b) \\ =lg (a(a+b)) \\ =\lg (a^2+ab)$<br /> <br /> ==Oppgave 6==<br /> ==Oppgave 7==<br /> ==Oppgave 8==<br /> \displaystyle 2x^{3}+6x^{2}=0<br /> <br /> ==Oppgave 9==<br /> <br /> ==Oppgave 10==</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12562 2014-02-22T21:05:58Z

<p>ThomasSkas: /* Oppgave 8 */</p> <hr /> <div>==DEL EN==<br /> <br /> ==Oppgave 1==<br /> <br /> $f(x)= 3x^2+3x+1 \qquad D_f = \R \\f'(x)= 6x-3 \\f´(2)= 6 \cdot 2 -3 = 9$<br /> <br /> ==Oppgave 2==<br /> <br /> <br /> ===a)===<br /> $x(x+5)-10 =4 \\ x^2+5x-14=0 \\ x= \frac{-5\pm \sqrt{25+56}}{2} \\x= -7 \vee x=2$<br /> <br /> ===b)===<br /> <br /> $10^{3x} -100000 =0 \\ 10^{3x} = 10^5 \\ 3x=5 \\ x=\frac 53$<br /> <br /> ==Oppgave 3==<br /> ==Oppgave 4==<br /> <br /> <br /> ===a)===<br /> $v=v_0+at \\t = \frac{v-v_0}{a}$<br /> <br /> ===b)===<br /> $t = \frac{v-v_0}{a} \\<br /> t = \frac{25-1}{3}=8 $<br /> <br /> ==Oppgave 5==<br /> <br /> ===a)===<br /> <br /> $ \frac{9^2a^2b^3}{(3ab^2)^2} \\ =\frac{3^4a^2b^3}{3^2a^2b^6} \\ = 3^{4-2} \cdot a^{2-2} \cdot b^{3-6} \\ = 3^2b^{-3} \\ = \frac{9}{b^3} $<br /> <br /> ===b)===<br /> <br /> $\lg (\frac{a^2}{b^2}) + \lg( \frac {b^2}{a}) + lg (a+b) \\ =\lg a^2 - \lg b^2 + \lg b^2 - \lg a + \lg (a+b) \\= 2\lg a - \lg a + \lg(a+b) \\ = \lg a + \lg (a+b) \\ =lg (a(a+b)) \\ =\lg (a^2+ab)$<br /> <br /> ==Oppgave 6==<br /> ==Oppgave 7==<br /> ==Oppgave 8==<br /> a) <br /> <br /> \displaystyle 2x^{3}+6x^{2}=0<br /> <br /> ==Oppgave 9==<br /> <br /> ==Oppgave 10==</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12561 2014-02-22T19:39:08Z

<p>ThomasSkas: /* Oppgave 8 */</p> <hr /> <div>==DEL EN==<br /> <br /> ==Oppgave 1==<br /> <br /> $f(x)= 3x^2+3x+1 \qquad D_f = \R \\f'(x)= 6x-3 \\f´(2)= 6 \cdot 2 -3 = 9$<br /> <br /> ==Oppgave 2==<br /> <br /> <br /> ===a)===<br /> $x(x+5)-10 =4 \\ x^2+5x-14=0 \\ x= \frac{-5\pm \sqrt{25+56}}{2} \\x= -7 \vee x=2$<br /> <br /> ===b)===<br /> <br /> $10^{3x} -100000 =0 \\ 10^{3x} = 10^5 \\ 3x=5 \\ x=\frac 53$<br /> <br /> ==Oppgave 3==<br /> ==Oppgave 4==<br /> <br /> <br /> ===a)===<br /> $v=v_0+at \\t = \frac{v-v_0}{a}$<br /> <br /> ===b)===<br /> $t = \frac{v-v_0}{a} \\<br /> t = \frac{25-1}{3}=8 $<br /> <br /> ==Oppgave 5==<br /> <br /> ===a)===<br /> <br /> $ \frac{9^2a^2b^3}{(3ab^2)^2} \\ =\frac{3^4a^2b^3}{3^2a^2b^6} \\ = 3^{4-2} \cdot a^{2-2} \cdot b^{3-6} \\ = 3^2b^{-3} \\ = \frac{9}{b^3} $<br /> <br /> ===b)===<br /> <br /> $\lg (\frac{a^2}{b^2}) + \lg( \frac {b^2}{a}) + lg (a+b) \\ =\lg a^2 - \lg b^2 + \lg b^2 - \lg a + \lg (a+b) \\= 2\lg a - \lg a + \lg(a+b) \\ = \lg a + \lg (a+b) \\ =lg (a(a+b)) \\ =\lg (a^2+ab)$<br /> <br /> ==Oppgave 6==<br /> ==Oppgave 7==<br /> ==Oppgave 8==<br /> a) f(x)=0<br /> <br /> 2x^{3}-6x^{2}=0<br /> <br /> ==Oppgave 9==<br /> <br /> ==Oppgave 10==</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12560 2014-02-22T19:36:40Z

<p>ThomasSkas: /* Oppgave 8 */</p> <hr /> <div>==DEL EN==<br /> <br /> ==Oppgave 1==<br /> <br /> $f(x)= 3x^2+3x+1 \qquad D_f = \R \\f'(x)= 6x-3 \\f´(2)= 6 \cdot 2 -3 = 9$<br /> <br /> ==Oppgave 2==<br /> <br /> <br /> ===a)===<br /> $x(x+5)-10 =4 \\ x^2+5x-14=0 \\ x= \frac{-5\pm \sqrt{25+56}}{2} \\x= -7 \vee x=2$<br /> <br /> ===b)===<br /> <br /> $10^{3x} -100000 =0 \\ 10^{3x} = 10^5 \\ 3x=5 \\ x=\frac 53$<br /> <br /> ==Oppgave 3==<br /> ==Oppgave 4==<br /> <br /> <br /> ===a)===<br /> $v=v_0+at \\t = \frac{v-v_0}{a}$<br /> <br /> ===b)===<br /> $t = \frac{v-v_0}{a} \\<br /> t = \frac{25-1}{3}=8 $<br /> <br /> ==Oppgave 5==<br /> <br /> ===a)===<br /> <br /> $ \frac{9^2a^2b^3}{(3ab^2)^2} \\ =\frac{3^4a^2b^3}{3^2a^2b^6} \\ = 3^{4-2} \cdot a^{2-2} \cdot b^{3-6} \\ = 3^2b^{-3} \\ = \frac{9}{b^3} $<br /> <br /> ===b)===<br /> <br /> $\lg (\frac{a^2}{b^2}) + \lg( \frac {b^2}{a}) + lg (a+b) \\ =\lg a^2 - \lg b^2 + \lg b^2 - \lg a + \lg (a+b) \\= 2\lg a - \lg a + \lg(a+b) \\ = \lg a + \lg (a+b) \\ =lg (a(a+b)) \\ =\lg (a^2+ab)$<br /> <br /> ==Oppgave 6==<br /> ==Oppgave 7==<br /> ==Oppgave 8==<br /> a) f(x)=0<br /> <br /> \displaystyle 2x^{3}-6x^{2}=0<br /> <br /> ==Oppgave 9==<br /> <br /> ==Oppgave 10==</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12559 2014-02-22T19:33:08Z

<p>ThomasSkas: /* Oppgave 8 */</p> <hr /> <div>==DEL EN==<br /> <br /> ==Oppgave 1==<br /> <br /> $f(x)= 3x^2+3x+1 \qquad D_f = \R \\f'(x)= 6x-3 \\f´(2)= 6 \cdot 2 -3 = 9$<br /> <br /> ==Oppgave 2==<br /> <br /> <br /> ===a)===<br /> $x(x+5)-10 =4 \\ x^2+5x-14=0 \\ x= \frac{-5\pm \sqrt{25+56}}{2} \\x= -7 \vee x=2$<br /> <br /> ===b)===<br /> <br /> $10^{3x} -100000 =0 \\ 10^{3x} = 10^5 \\ 3x=5 \\ x=\frac 53$<br /> <br /> ==Oppgave 3==<br /> ==Oppgave 4==<br /> <br /> <br /> ===a)===<br /> $v=v_0+at \\t = \frac{v-v_0}{a}$<br /> <br /> ===b)===<br /> $t = \frac{v-v_0}{a} \\<br /> t = \frac{25-1}{3}=8 $<br /> <br /> ==Oppgave 5==<br /> <br /> ===a)===<br /> <br /> $ \frac{9^2a^2b^3}{(3ab^2)^2} \\ =\frac{3^4a^2b^3}{3^2a^2b^6} \\ = 3^{4-2} \cdot a^{2-2} \cdot b^{3-6} \\ = 3^2b^{-3} \\ = \frac{9}{b^3} $<br /> <br /> ===b)===<br /> <br /> $\lg (\frac{a^2}{b^2}) + \lg( \frac {b^2}{a}) + lg (a+b) \\ =\lg a^2 - \lg b^2 + \lg b^2 - \lg a + \lg (a+b) \\= 2\lg a - \lg a + \lg(a+b) \\ = \lg a + \lg (a+b) \\ =lg (a(a+b)) \\ =\lg (a^2+ab)$<br /> <br /> ==Oppgave 6==<br /> ==Oppgave 7==<br /> ==Oppgave 8==<br /> a) f(x)=0<br /> <br /> ==Oppgave 9==<br /> <br /> ==Oppgave 10==</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12531 2014-02-19T15:11:10Z

<p>ThomasSkas: </p> <hr /> <div>Løsning<br /> <br /> Oppgave 1<br /> <br /> [tex]f'(x)= 6x-3[/tex]</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12530 2014-02-19T15:09:27Z

<p>ThomasSkas: </p> <hr /> <div>Løsning<br /> <br /> Oppgave 1<br /> <br /> f'(x) = 6x-3<br /> <br /> f'(2) = 6*2-3 = 9</div>

ThomasSkas https://matematikk.net/w/index.php?title=S1_2013_h%C3%B8st_L%C3%98SNING&diff=12529 2014-02-19T15:07:30Z

<p>ThomasSkas: Ny side: Løsning</p> <hr /> <div>Løsning</div>

ThomasSkas