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3 years ago

Find the missing length for a right triangle with hypotenuse of 18 and a leg length of 6

Mathematics

2 answers:

Alla [95]3 years ago

8 0

16.9705624748447714
if you want it rounded its 16.97 or 16.9
thanks for letting me answer and god bless <3

yaroslaw [1]3 years ago

4 0

$\sqrt{18^2-6^2}$
$\sqrt{256-36} = \sqrt{220} = 2 \sqrt{55}$

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I am making quilts out of my grandma's square handkerchief collection. I have

Yuki888 [10]

Answer:

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Step-by-step explanation:

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3 years ago

Multiply (-2xy+z)(5x+yz).

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3 years ago

..vertex: (4, -1)<br> f(3) = -6

s344n2d4d5 [400]

Answer:

$\huge\boxed{f(x)=-5(x-4)^2-1=-5x^2+40x-81}$

Step-by-step explanation:

The vertex form of an equation of a parabola:

$f(x)=a(x-h)^2+k$

$(h;\ k)$ - vertex

We have

vertex $(4;\ -1)\to h=4;\ k=-1$

$f(3)=-6$

Therefore we have:

$f(x)=a(x-4)^2+(-1)=a(x-4)^2-1$

Substitute $x=3$ and $f(3)=-6$

$a(3-4)^2-1=-6\qquad|\text{add 1 to both sides}\\\\a(-1)^2=-6+1\\\\a(1)=-5\\\\a=-5$

Finally:

$f(x)=-5(x-4)^2-1=-5(x^2-2(x)(4)+4^2)-1\\\\=-5(x^2-8x+16)-1=-5x^2+(-5)(-8x)+(-5)(16)-1\\\\=-5x^2+40x-80-1=-5x^2+40x-81$

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GREYUIT [131]

Answer:

I think it is 5a+2ab+12

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3 years ago