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3 years ago

NO LINKS, HELP IM GIVING BRAINLIEST TO CORRECT ANSWER

Mathematics

1 answer:

julsineya [31]3 years ago

3 0

Answer:

100

Step-by-step explanation:

$8.(33) * \sqrt{144} = 8.(3) + \sqrt{144} = 8 \frac{1}{3} * 12 = \frac{25}{3} * 12 = 25 * 4 = 100$

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What is the solution of the system ?Use substitution.<br> Y= -3x<br> X + y = -4

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I hope this helps you

x-3x= -4

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y= -3.2

y= -6

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F(x) =x^2-5x write down the value of ff(2).

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3 years ago

Question 9

Alenkasestr [34]

Answer:

$PX=4\\PY=4\sqrt{3}$

Step-by-step explanation:

Let $\angle PYZ=Q$

$\Rightarrow \angle XYP=90-\angle PYZ\\\angle XYP=90-\theta$

In the $\Delta PYZ$

$\tan\theta =\frac{opposite}{adjacent}=\frac{PZ}{PY}......(1)$

In the $\Delta YPX$

$\tan(90-\theta)=\frac{opposite}{adjacent}=\frac{PX}{PY}$

$\cot=\frac{PX}{PY}....(2)$

eqn(1) $\times$ eqn(2)

$\tan\times\cot\theta=\frac{PZ}{PY}\times\frac{PX}{PY}\\\\1=\frac{PZ\times PX}{PY^2}\ \ (as\ tan\theta\ =\frac{1}{\cot\theta})\\\\PY^2=PZ\times PX\\PY^2=12PX.......(3)$

Now in $\Delta XYP$

use Pythagorean theorem

$XY^2=PY^2+PX^2\\8^2=PY^2+PX^2\ \ (as\ XY\ =8)\\PX^2+PY^2=64\\\\PY^2=12PX\ \ (eqn(3)\\\\\Rightarrow PX^2+12PX=64\\\\PX^2+12PX-64=0\\\\PX^2+16-4PX-64=0\\\\(PX+16)(PX-4)=0\\\\PX=4,\ -16$

Length can not be negative

Hence $PX=4$

$PY^2=12PX=12\times4=48\\\\PY=\sqrt{48}=4\sqrt{3}$

Hence $PX=4,\ PY=4\sqrt{3}$

8 0

3 years ago