URGENT!!! plz help me with these two

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer:

hope that helps!!

Step-by-step explanation:

the first one is A

the second one is (2,-1)

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Geometry math question

Brut [27]

Look at the picture.

Let |AE| = |AD| = b

We have a proportion:

$\dfrac{x}{13}{y}=\dfrac{19}{y+a}=\dfrac{x}{y+2a}$

Solve for y from first proportion

$\dfrac{13}{y}=\dfrac{19}{y+a}\ \ \ |\text{cross multiply}\\\\19y=13(y+a)\\\\19y=13y+13a\ \ \ \ |-13y\\\\6y=13a\ \ \ |:6\\\\y=\dfrac{13}{6}a$

Substitute to the second proportion

$\dfrac{19}{y+a}=\dfrac{x}{y+2a}\\\downarrow\\\dfrac{19}{\frac{13}{6}a+a}=\dfrac{x}{\frac{13}{6}a+2a}\\\\\dfrac{19}{\frac{13}{6}a+\frac{6}{6}a}=\dfrac{x}{\frac{13}{6}a+\frac{12}{6}a}\\\\\dfrac{19}{\frac{19}{6}a}=\dfrac{x}{\frac{25}{6}a}\ \ \ \ |\dot a\neq0\\\\\dfrac{19}{\frac{19}{6}}=\dfrac{x}{\frac{25}{6}}\\\\19\cdot\dfrac{6}{19}=x\cdot\dfrac{6}{25}\\\\6=\dfrac{6x}{25}\ \ \ |\cdot25\\\\6x=6\cdot25\ \ \ |:6\\\\x=25$