Can someone help me solve this?

Mathematics

1 answer:

katen-ka-za [31]3 years ago

5 0

See the pic for the answer. You just multiply each term on the left by each term on the top. The coefficient is just the number in front of an unknown.

You might be interested in

If CDE ~ GDF, find ED

qaws [65]

Answer:

Step-by-step explanation:

$\triangle CDE \sim \triangle GDF.. (given) \\\\\therefore \frac{CD}{GD} =\frac{DE}{DF}.. (csst) \\\\\therefore \frac{15}{x+3} =\frac{3x+1}{4}\\\\ \therefore \: 15 \times 4 = (x + 3)(3x + 1) \\ \\ \therefore \: 60 = 3 {x}^{2} + x + 9x + 3 \\ \\ \therefore 3 {x}^{2} + 10x + 3 - 60 = 0 \\ \therefore 3 {x}^{2} + 10x - 57 = 0 \\ \therefore 3 {x}^{2} + 19x - 9x - 57 = 0 \\ \therefore \: x(3x + 19) - 3(3x + 19) = 0 \\\therefore \: (3x + 19)(x - 3) = 0 \\ \therefore \: 3x + 19 = 0 \: \: or \: \: x - 3 = 0 \\ \therefore \: x = - \frac{19}{3} \: \: or \: \: x = 3 \\ \because \: x \: can \: not \: be \: - ve \\ \therefore \: x = 3 \\ ED = 3x + 1 = 3 \times 3 + 1 \\ \huge \red{ \boxed{ ED= 10}}$