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3 years ago

Name the triangle whose angles are all less than 90.

1 answer:

3 0

It's acute triangle.
obtuse is where all of the angles are over 90
and right triangle is when one of the angles are exactly 90, and isosceles triangle is where two sides has the same length.

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Meghan Trainor spent $180 to start a dog walking business and earns $80 for each dog she walks Nicki Minaj didn't spend anything

REY [17]

Answer:

Nicki will have to walk 13 dogs to get the same amount of money as Meghan

Pls mark me brainliest.

4 0

2 years ago

Aplumber charges a $75 flat fee for jobs lasting up to an hour and $30 for each hour of labor after the first hour. Which

tamaranim1 [39]

<h3><u>Answer</u>:- </h3>

$\longrightarrow \sf 75+ 30(h-1)$

<h3><u>Solution:-</u></h3>

<u>Given </u><u>that </u><u>:</u><u>-</u>

job last for h hours
Cost of job lasting more than 1 hour = $30
Cost for the first hour = $75

$\leadsto$ Remaining hours for which the job will last after first hour= (h-1)

<u>Therefore</u>,

Cost for subsequent hours = 30×(h-1)

= 30(h-1)

Total Cost for job = Cost for 1st hour + Cost for subsequent hours

$\quad\quad\quad\quad\quad\quad\quad\quad\boxed{\sf{= 75+30(h-1)}}$

7 0

2 years ago

Help please asappppppppp

Aleonysh [2.5K]

Answer:

First Equation: 6a-23

Second Equation: 5a-20

Step-by-step explanation:

So, I think you want me to answer part 2...

For the first equation, we can see that the denominator stays the exact same, meaning the numerator will also not change.

For the second equation, we can see the denominator changed from a-3 to 5a-15. The denominator was simply multiplied by 5. Since the denominator was multiplied by five we must multiply the numerator by five as well.

5(a-4)

5a-20

Hope this helps :)

3 0

2 years ago

What is the volume of this cone?

Vika [28.1K]

Answer:

The answer is A 41.9 feet.

4 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago