All categories

3 years ago

Given the following figure. Which of the following is congruent with the figure above

Mathematics

1 answer:

tigry1 [53]3 years ago

3 0

F and E are congruent with the figure above

You might be interested in

A line with a slope of 3 passes through (2,5). Which choice is an equation of this line? Please help

jek_recluse [69]

Answer:

Top right.

Step-by-step explanation:

So we want a line with a slope of 3 and passes through (2,5).

To do so, we can use the point-slope form.

The point-slope form is:

$y-y_1=m(x-x_1)$

m is the slope and x₁ and y₁ is an ordered pair.

Thus, let m be 3, y₁ be 5, and x₁ be 2. Thus:

$y-5=3(x-2)$

Our answer is the top right :)

3 0

3 years ago

Read 2 more answers

For altitudes up to 36,000 feet, the relationship between ground temperature and atmospheric

Verizon [17]

Answer:

The equation for a is $a=-\frac{2000}{7}*(t-g)$

The altitute is 101,428.57 feet

Step-by-step explanation:

You know that the relationship between ground temperature and atmospheric temperature can be described by the formula

t = -0.0035a +g

where:

t is the atmospheric temperature in degrees Fahrenheit
a is the altitude, in feet, at which the atmospheric temperature is measured
g is the ground temperature in degrees Fahrenheit.

Solving the equation for a:

-0.0035a +g=t

-0.0035a= t - g

$a=\frac{t-g}{-0.0035}$

$a=-\frac{2000}{7}*(t-g)$

The equation for a is $a=-\frac{2000}{7}*(t-g)$

If the atmospheric temperature is -305 °F and the ground temperature is 50 °F, then t= -305 °F and g= 50 °F

Replacing in the equation for a you get:

$a=-\frac{2000}{7}*(-305-50)$

$a=-\frac{2000}{7}*(-355)$

a= 101,428.57

The altitute is 101,428.57 feet

4 0

3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical cap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago

Find the distance from the point (6,10) to the line y=-x to the nearest tenth

ValentinkaMS [17]

points (6,10)
y = -x
x + y = 0

distance = lax1 + by1 + cl/√(a^2 + b^2)
= l1(6) + 1(10) + 0l/√(1^2 + 1^2)
= l6 + 10 + 0l/√(1 + 1)
= l16l/√2
= 16/√2
= 8 .2/√2
= 8 . √2.√2/√2
= 8√2

4 0

3 years ago

Can someone please explain how to do these?

Pani-rosa [81]

Answer:

First question answer: The limit is 69

Second question answer: The limit is 5

Step-by-step explanation:

For the first limit, plug in $x=8$ in the expression $(9x-3)$ , that's the answer for linear equations and limits.

So we have:

$9x-3\\9(8)-3\\72-3\\69$

The answer is 69

For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value $x=1$ into the simplified expression to get the correct answer. Shown below:

$\frac{x^2+8x-9}{x^2-1}\\\frac{(x+9)(x-1)}{(x-1)(x+1)}\\\frac{x+9}{x+1}$

Now putting 1 in $x$ gives us the limit:

$\frac{x+9}{x+1}\\\frac{1+9}{1+1}=\frac{10}{2}=5$

So the answer is 5

3 0

3 years ago