Triangle P Q R is shown. Angle Q P R is a right angle. The length of Q P is 8 StartRoot 3 EndRoot and the length of P R is 8.

Can any one help me 11 points no lnks pls. Just help with any of these

zvonat [6]

Answer:

5. m<1 = 114 degrees

m<2 = 66 degrees

6. 2x + 3x + 4x = 9x = 180, x = 20

7. 3

8. 1/3

Hope that helps!

-scsb

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

HELP ME TVT I beg you (answer by yes or no)

aliya0001 [1]

Answer:

1:yes 2:no 3:no 4:yes

Step-by-step explanation:

3*2=6 for the numerators

4*5=20 for the denominators

you can simplify this to make it 3/10 ✔️

2*5-10 for the numerators

6*5=30 for the denominators

you can simplify this to make it 1/3 ❌

1*1=1 for the numerators

10*5 for the denominators

you can turn this into 1/50 ❌

1*3=3 for the numerators

2*5=10 for the denominators

you can turn this into 3/10 ✔️

hope this helps! have a nice day! :)

5 0

2 years ago

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

nataly862011 [7]

Let (x, y, z) be a point on the plane in the first octant. The box formed by this point has volume xyz, and you want to maximize this subject to the equation of the plane.

Use the method of Lagrange multipliers: the Lagrangian is

L(x, y, z) = xyz - λ (x + 2y + 3z - 6)

Find its critical points:

∂L/∂x = yz - λ = 0

∂L/∂y = xz - 2λ = 0

∂L/∂z = xy - 3λ = 0

∂L/∂λ = -(x + 2y + 3z - 6) = 0

Solving the first three equations for λ gives

λ = yz = xz/2 = xy/3

Solve these equations for y and z :

• yz = xz/2 => y = x/2 => 2y = x

• yz = xy/3 => z = x/3 => 3z = x

Substitute these solutions into the last equation and solve for x, then again for y and z :

x + 2y + 3z - 6 = 3x - 6 = 0 => 3x = 6 => x = 2, y = 1, z = 2/3

At this critical point, the maximum volume is

xyz = 2*1*2/3 = 4/3

4 0

3 years ago

A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then s

Bas_tet [7]

Answer:

Step-by-step explanation:

$m=\frac{3}{32}\\ \gamma =0\text{ damping constant.}\\ k=\frac{3}{\frac{1}{4}}=12\\ F(t)=0\\ \text{so equation of motion is }\\ mu''+\gamma u'+ku=F(t)\\ \frac{3u''}{32}+12u=0\\ u''+128u=0\\ \text{and initial conditions are}\\ u(0)=-\frac{1}{12}\\ u'(0)=2\\ u''+128u=0\\ \text{characterstic equation is}\\ r^2+128=0\\ \text{roots are }\\ r=\pm 8\sqrt{2}i\\ \text{therefore general solution is}\\ u(t)=A\cos8\sqrt{2}t+Bsin8\sqrt{2}t\\ \text{using initial conditions we get}\\ A=-\frac{1}{12}\\ B=\frac{\sqrt{2}}{8}\\ \text{therefore solution is }\\ u(t)=-\frac{1}{12}\cos8\sqrt{2}t+\frac{\sqrt{2}}{8}\sin8\sqrt{2}t\\
{hence}\\R=\sqrt{\frac{11}{288}}\\\\\sigma=\pi - \tan^{-1}\frac{1}{\sqrt{2}}\\\\\omega_0=8\sqrt{2}\\T=\frac{4}{4\sqrt{2}}$