Please help meh ? :(

Helppppppppppp please and expalin if you can

Klio2033 [76]

The angle subtended by a diameter is always right. That's Thales' Theorem, one of the first in geometry. We conclude y=90°.

The angle subtended by an arc is half the arc length, so z=50°.

The angles in a triangle add to 180°, so x = 180° - y - z = 40°.

Answer: x=40°, y=90°, z=50°

8 0

3 years ago

Which of the following is the point slope form of the linear equation with the following solutions?

Nana76 [90]

Answer:

Third choice: y + 7 = -2(x - 3)

Step-by-step explanation:

One of the two given points is (3, -7): (h, k), which would be the given point in the point-slope form of the equation of a straight line:

y - k = m(x - h)

If we replace h with 3 and k with -7, we get

y + 7 = m(x - 3).

Next, let's calculate the slope, m: as we go from (2, -5) to (3, -7), x increases by 1 and y decreases by 2. Hence, the slope is m = rise / run = -2/1, or -2.

Thus, the third answer choice is the correct one: y + 7 = -2(x - 3).

8 0

4 years ago

Trapezoid WXYZ has vertices W(−1, 2) , X(2, 2) , Y(3, −1) , and Z(−3, −1) .

Scorpion4ik [409]

She is right its ( is not) just took the test

5 0

3 years ago

Read 2 more answers

Find the directional derivative of the function at the given point in the direction of the vector v. h(r, s, t) = ln(3r + 6s + 9

vodka [1.7K]

Answer:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})$

Step-by-step explanation:

First, let’s check to see if the direction vector is a unit vector:

$||v||=\sqrt{(14)^{2} +(42)^{2} +(21)^{2} } =\sqrt{2401} =49$

It’s not a unit vector. therefore let's divide the vector by its magnitude in order to convert it into a unit vector:

$\overrightarrow{\rm v}=(\frac{14}{49}\overrightarrow{\rm i})+(\frac{42}{49}\overrightarrow{\rm j})+(\frac{21}{49}\overrightarrow{\rm k})= (\frac{2}{7}\overrightarrow{\rm i})+(\frac{6}{7}\overrightarrow{\rm j})+(\frac{3}{7}\overrightarrow{\rm k})$

Now, the directional derivative is given by:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=\frac{\partial h(r,s,t)}{\partial r}\overrightarrow{\rm i} + \frac{\partial h(r,s,t)}{\partial s}\overrightarrow{\rm j} + \frac{\partial h(r,s,t)}{\partial t}\overrightarrow{\rm k}$

So let's calculate the partial derivates:

$\frac{\partial }{\partial r} ln(3r+6s+9t)=\frac{3}{3r+6s+9t}=\frac{1}{r+2s+3t}$

$\frac{\partial }{\partial s} ln(3r+6s+9t)=\frac{6}{3r+6s+9t}=\frac{2}{r+2s+3t}$

$\frac{\partial }{\partial t} ln(3r+6s+9t)=\frac{9}{3r+6s+9t}=\frac{3}{r+2s+3t}$

Therefore:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})$

3 0

4 years ago

Mia ran around 0.10 of the school track. Sam ran around 0.5 of the playground. Sam says he ran farther. With the information giv

dybincka [34]

No sam is not correct because Mia ran .10 and Sam ran .05 And although .05 may seem bigger .10 is more expressed That is the answer I think

3 0

4 years ago

Read 2 more answers