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yawa3891 [41]
3 years ago
5

Two angles are supplementary, and one angle is one-fifth the other. Find the measure of the two angles.

Mathematics
1 answer:
choli [55]3 years ago
5 0

:Two Angles are Supplementary when they add up to 180 degrees. Notice that together they make a straight angle.

and if that doesn't help

:There are in an angle of one fifth of the supplementary angle. Supplementary angle decides the angle to always be and the complementary angle is always (i.e., right angle)

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If P is the incenter of JKL, find each measure
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Answer:

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X²=14x-45 step by step please
dolphi86 [110]

Answer:

X=9 and X=5

Step-by-step explanation:

So you want to get one side to equal zero so you can factor and find x. I recommend subtracting 14x and adding 45 to both sides so you don't have to deal with a negative quadratic.

X^2 - 14X +45 = 0

Now you can factor. Your looking for two factors that equal 45 and add to -14

All factors of 45:

1 and 45, 3 and 15, 5 and 9, -1 and -45,

-3 and -15, -5 and -9

So out of those combinations -5 and -9 both multiply to 45 and add up to -14 so these are our common factors

(X-5)(X-9)=0

X-5=0 add 5 to both sides X=5

X-9=0 add 9 to both sides X=9

5 0
3 years ago
Aşağıda verilen şıklarda istenenleri bulunuz (NOT: sqrt( ): karekök, örneğin sqrt(4)=2 dir, R+: pozitif reel sayılar gösterimi).
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3 years ago
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<u>Step-by-step explanation:</u>

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4 0
3 years ago
Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your
BaLLatris [955]

f(x,y)=\dfrac{y^3}x

a. The gradient is

\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath

\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}

b. The gradient at point P(1, 2) is

\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}

c. The derivative of f at P in the direction of \vec u is

D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}

It looks like

\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath

so that

\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2

Then

D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}

\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}

7 0
3 years ago
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