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3 years ago

10

see i don't mind giving my points away..Can anyone help me solve this though for a brainlest and extra points?

2 answers:

Oksana_A [137]3 years ago

7 0

Answer:

17sqrt2

Step-by-step explanation:

45 45 90 triangles side: hypotenuse a:a*sqrt2

17*sqrt2=17sqrt2

victus00 [196]3 years ago

6 0

Answer:

a= $17\sqrt{2}$

Step-by-step explanation:

As we have 2 of the angles, we can find the third. The measure of the third angle is 45 degrees.

As this is the same measure as the other, that means that the side length will be the same.

Now was have two side lengths of 17 in a right triangle, so we can use the Pythagorean theorem to find a.

Recall that the pythagorean theorem states: $a^2+b^2=c^2$

In this case, a is 17, b is 17, and c is a

Knowing this, we can input our value into this formula and solve for a.

$17^2+17^2=a^2\\\\289+289=a^2\\\\a=\sqrt{578} \\\\a=17\sqrt{2}$

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cluponka [151]

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2 years ago

Still havent recieved the CORRECT ANSWERS... u will be the brainliest!

yKpoI14uk [10]

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3 years ago

You earn $21.00 per hour at your job. If you get a 9% raise at the end of each year, what will your hourly rate, h, be after 5 y

NikAS [45]

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3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

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3 years ago

24 POINTS!!!!!!!!!!!!!!!!!!

4vir4ik [10]

The answer is the angles are neither complementary nor supplementary, because complementary is to add up to 90° and supplementary is to add up to 180°, and these angles add up to 169°. I hope this helps!

3 0

2 years ago

Read 2 more answers