Ask question

Ask question

All categories

2 years ago

6

For an outdoor track meet to be cancelled, the temperature, t, outside must be colder than 35 degrees. Complete the following in

equality by dragging the correct inequality symbol to the box.

2 answers:

qwelly [4]2 years ago

7 0

Answer: I'm pretty sure it would be -7º Celcius for it to be cancelled. Hope it helped!

Scilla [17]2 years ago

7 0

Answer:

byvyv6v6v6v6v

Step-by-step explanation:

byvyv6v6v6

You might be interested in

The percent decrease from 12 to 9 is equal to the percent decrease from 40 to what number

Zina [86]

(12 - 9) / 12 = (40 - x) / 40
3/12 = (40 - x) / 40
40(1/4) = 40 - x
10 = 40 - x
10 - 40 = -x
-30 = -x
30 = x <=== percent decrease from 40 to 30

8 0

1 year ago

What is 6 divided by 612

kondaur [170]

612/6=102
check: 6x102=612

Hope this helped! :))

8 0

1 year ago

Read 2 more answers

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

1 year ago

Seth is using the figure shown below to prove the Pythagorean Theorem using triangle similarity: In the given triangle PQR, angl

Naily [24]

Answer:

Part A: $\triangle RPQ \sim \triangle RSP$

Part B. All angles are same, so the triangles are similar.

Part C. RP = 8

Step-by-step explanation:

We are given a right angled triangle $\triangle RPQ$ with $\angle P = 90^\circ$ .

PS is perpendicular to the hypotenuse RQ of $\triangle RPQ$ and S lies on RQ.

Part A:

To identify the pair of similar triangles.

$\triangle RPQ \sim \triangle RSP$ .

Part B:

To identify the type of similarity.

Kindly refer to the image attached in the answer area.

Let us consider the triangles $\triangle RPQ \ and\ \triangle RSP$ .

$\angle RSP =\angle RPQ =90^\circ$

Also, $\angle R$ is common to both the triangles under consideration.

Now, we can see that two angles of two triangles are equal.

So, third angle of the two triangles will also be same.

i.e. All three angles of two triangles $\triangle RPQ \ and\ \triangle RSP$ are equal to each other.

So, by A-A-A (Angle - Angle - Angle) similarity, we can say that $\triangle RPQ \sim \triangle RSP$ .

Part C:

RS = 4

RQ = 16, Find RP.

There is one property of similar triangles that:

The ratio of corresponding sides of two similar triangles is equal.

i.e.

$\dfrac{RS}{RP} = \dfrac{RP}{RQ}\\\Rightarrow RP ^2 = RS \times RQ\\\Rightarrow RP ^2 = 4 \times 16\\\Rightarrow RP ^2 = 64\\\Rightarrow \bold{RP = 8\ units}$

5 0

2 years ago

Noelle stands at the edge of a cliff and drops a rock. The height of the rock is given by the function f(x)-4.9x2 + 15, where c

Oxana [17]

A graphing calculator shows the rocks are at the same height 1.5 seconds after they are released.

That height is 3.975 meters.

_____
f(x) = g(x)
-4.9x^2 +15 = -4.9x^2 +10x
15 = 10x . . . . . . . . . . . . . . . . . . add 4.9x^2
1.5 = x . . . . . . . . . . . . . . . . . . . divide by 10
f(1.5) = -4.9*2.25 +15 = 3.975

3 0

2 years ago