Need help ASAP please

Answer number 1 please.

Minchanka [31]

$\sf {a}^{2} = {13}^{2} - {5}^{2}$

Formula :

Base²= Hypotenuse² - Perpendicular ²

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$\hookrightarrow\sf {a}^{2} = {13}^{2} - {5}^{2}$

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$\hookrightarrow\sf {a}^{2} =169 - 25$

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$\hookrightarrow\sf {a}^{2} =144$

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$\hookrightarrow\sf {a} = \sqrt{144}$

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$\hookrightarrow\sf {a} = \sqrt{12 \times 12}$

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$\hookrightarrow\sf {a} = 12$

Remember the a² in formula has nothing to do with the a we have to find. :)

6 0

2 years ago

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Factor completely. y2+6y+8

zloy xaker [14]

If you can't see the picture the answer is: (y+4)(y+2).

Hope this helps, and May the Force Be With You!
-Jabba

4 0

3 years ago

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83 Tee (CsT Sequences and Functions

NISA [10]

Answer:

The common difference is same = d = -9

Therefore, the data represent a linear function.

Step-by-step explanation:

Given the table

x y

1 4

2 -5

3 -14

4 -23

5 -32

Finding the common difference between all the adjacent terms of y-values

d = -5 - 4 = -6,

d = -14 - (-5) = -14+5 = -9

d = -23 - (-14) = -23 + 14 = -9

d = -32 - (-23) = -32 + 23 = -9

It is clear that the common difference between all the adjacent terms is same.

Thus,

d = -9

We know that when y varies directly with x, the function is a linear function.

Here, it is clear that each x value varies 1 unit, and each y value varies -9 units.

i.e. The common difference is same = d = -9

Therefore, the data represent a linear function.

3 0

2 years ago

Help thanks//////////??????????

3241004551 [841]

Answer:

Step-by-step explanation:

At the point where two line cut across, the vertically opposite angles formed are equal. This means that

2x degrees = 78 degrees

Dividing the left hand side and the right hand side of the equation by 2, it becomes

2x/2 = 78/2

x = 39

Checking,

The sum of the angles on a straight line is 180 degrees.

180 - 78 = 102 degrees

2x + 102 = 180

2x = 180 - 102 = 78

x = 78/2 = 39

8 0

3 years ago

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Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
$L_\lambda=x+y+z-5=0$

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.

4 0

3 years ago