Recent content by zelda1850

so my answer would be -4 + 2\sqrt{5k} - 2\sqrt{5} + 5\sqrt{k}

can i add 2\sqrt{5k} + 5\sqrt{k} or it doesn't work

oh how can i correct it would it be sqaure root 25k?

that was the right question so can someone tell me what i did wrong?

im not sure what the like terms would be

ok can someone check if I am doing these questions correctly now? 1) (\sqrt{2a} -5)(7\sqrt{2a} - 5) \sqrt{2a} * 7\sqrt{2a} = 7 * 2a = 14a \sqrt{2a} * 5 = 5\sqrt{2a} 5 * 7\sqrt{2a} = 35\sqrt{2a} 5 * 5 = 25 = 35\sqrt{2a} - 5\sqrt{2a} = 30\sqrt{2a} = 14a + 30\sqrt{2a} - 25 2)...

tried to solve the problem with foil method but i got confuse can someone help 1) (-7 + \sqrt{3x})( 4+ \sqrt{3x}) -7 * 4 = -28 -7 * \sqrt{3x} = -7\sqrt{3x} \sqrt{3x} * 4 = 4\sqrt{3x} \sqrt{3x} * \sqrt{3x} - does this equal \sqrt{3x} square or is it just 3x or leave it \sqrt{3x} this is what...

ok so i was missing the little number am i doing question 2 correctly now? [SIZE="4"]-3\sqrt[3]{-3} + [SIZE="4"]2\sqrt[3]{162} + [SIZE="4"]3\sqrt[3]{81} [SIZE="4"]-3\sqrt[3]{-3} = [SIZE="4"]-3\sqrt[3]{3} [SIZE="4"]2\sqrt[3]{162} = [SIZE="4"]6\sqrt[3]{6} [SIZE="4"]3\sqrt[3]{81} =...

im not sure if I am doing these questions correctly can someone check it 1) [SIZE="4"]-3\sqrt[6]{3} - [SIZE="4"]2\sqrt[3]{192} - \sqrt[6]{320} [SIZE="4"]-3\sqrt[6]{3} = [SIZE="4"]-3\sqrt[6]{3} [SIZE="4"]2\sqrt[3]{192} = [SIZE="4"]6\sqrt[6]{3} \sqrt[6]{320} = [SIZE="4"]2\sqrt[6]{5}...

can u tell me how i can fix number 11 and number 7?

anyone mind checking my homework and help me with the problems i got wrong 1)-5\sqrt{3} - 3\sqrt{3} = -8\sqrt{3} 2)2\sqrt{8} - \sqrt{8} = 4\sqrt{2} - 2\sqrt{2} = 2\sqrt{2} 3)-4\sqrt{6} - \sqrt{6} = -4\sqrt{6} 4)-3\sqrt{5} + 2\sqrt{5} = \sqrt{5} 5)-3\sqrt{27} - 3\sqrt{27} -...

Homework Statement the magnitude of the electric field strength between two oppositly charged parallel metal plate is 2.0 x 10 3 Newtons per coulomb. point p is located midway between the plates. an electron is located at point p between the plates. caculate the magnitude of the force...

ok i got 12 r for the resistor using that equation

oh here a picture of the diagram for the problem

Homework Statement a 3.0 ohm resistor an unkown resistor r and two ammeters a1 and a2 are connected with a 12 volt source ammeter a2 reads a current of 5.0 amperes 1) detemine the equivalent resistance of the circuit 2) caculate the current measure of ammeter a 1 3) caculate the...