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3 years ago

Please help :(( ive been stuck for a while and im super late

Mathematics

1 answer:

Degger [83]3 years ago

3 0

Answer:

-5^11

Step-by-step explanation:

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The questions are below:

forsale [732]

Answer:

See below in BOLD.

Step-by-step explanation:

The first part:

Applying the Pythagoras theorem to both triangles.

The one on the left:

13^2 = 12^2 + x^2

x^2 = 169 - 144 = 25

x = 5 units.

The one on the right:

x^2 = 5^2 + 12^2 = 169

x = 13 units.

(a) So the 2 shorter sides in the triangles have the same length.

(b) The 3 sides of each triangle have the same length so they are congruent by SSS.

The Second Part.

(a) F. They could be congruent because the angles are all equal but we don't know anything about the lengths of the sides.

(b). C. Congruent. by ASA. The angles at each side of the line of length 9.2 are equal in measure.

8 0

4 years ago

Identify the slope and y-intercept of the line whose equation is given. Write the y-intercept as an ordered pair. (Remember the

xeze [42]

The answer is c. hope this helps.

6 0

3 years ago

Read 2 more answers

The equations that must be solved for maximum or minimum values of a differentiable function w=f(x,y,z) subject to two constrai

sammy [17]

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)$

with partial derivatives (set equal to 0)

$L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0$

$L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0$

$L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0$

$L_\lambda=4x^2+4y^2-z^2=0$

$L_\mu=2x+4z-2=0\implies x+2z=1$

Case 1: If $y=0$ , then

$4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|$

Then

$x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23$

$\implies x=\dfrac15\text{ or }x=-\dfrac13$

So we have two critical points, $\left(\dfrac15,0,\dfrac25\right)$ and $\left(-\dfrac13,0,\dfrac23\right)$

Case 2: If $\lambda=-\dfrac14$ , then in the first equation we get

$x(1+4\lambda)+\mu=\mu=0$

and from the third equation,

$z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0$

Then

$x+2z=1\implies x=1$

$4x^2+4y^2-z^2=0\implies1+y^2=0$

but there are no real solutions for $y$ , so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

$f\left(\dfrac15,0,\dfrac25\right)=\dfrac15$ (min)

and

$f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59$ (max)

5 0

4 years ago

Judith has already taken 93 quizzes during past quarters, and she expects to have 5 quizzes

lara [203]

Answer:

103

Step-by-step explanation:

93 + 5 + 5 = 103

6 0

3 years ago

Rewrite In simplest radical form x^5/6/x^1/6

Nady [450]

I assume you mean:

[x^(5/6)]/[x^(1/6)]

The rule for dividing similar bases with different exponents is:

(a^b)/(a^c)=a^(b-c)

In this case we have:

x^(5/6-1/6)

x^(4/6)

x^(2/3)

3 0

3 years ago