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3 years ago

Solve the linear expression 5(-3.4k-7)+(3k+21)

Mathematics

1 answer:

Ksivusya [100]3 years ago

6 0

Our expression is below

$5(-3.4k-7)+(3k+21) = 0$

by solving it for the value of k

$(-17k-35)+(3k+21) = 0$

$-17k+3k-35+21 = 0$

$-14k-14 = 0$

$-14k = 14$

$k = \frac{14}{-14}$

$k = -1$

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.

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Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

AD+DB=3

x+DB = 3

DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

$BC^{2} = DB^{2} +CD^{2}$

$BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}$

$BC^{2} =(3-x)^{2} +2$

Since given AD=BC,let us plugin BC=x in above step.

$x^{2} =(3-x)^{2} +2$

$x^{2} =9-6x+x^{2} +2$

6x=11

x= $\frac{11}{6}$

Now we know AD=x= $\frac{11}{6}$ and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

$AC^{2} =AD^{2} +DC^{2}$

$AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}$

$AC^{2} =\frac{193}{36}$

$AC=\sqrt{\frac{193}{36} }$ = 2.315cm

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