3 years ago

Where do fractions go on a number line

Mathematics

1 answer:

MatroZZZ [7]3 years ago

8 0

Answer:

fractions marked on number lines that span from 0 to 1.

Step-by-step explanation:

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Mia works at a job earning $4.75 per hour. How many hours should she work to earn $124.00?

labwork [276]

Mia would have to work 26.1 hours (approx. 26) to earn $124.00

4.75x=124

divide 4.75 by 4.75 and divide 124 by 4.75

x=26.1

8 0

3 years ago

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If the measure of angle 2 is 3x, and the measure of 1 is 5x - 12 what is the measure of angle 9 ?

Mariana [72]

3x+5x-12=180
8x=180+12
8x=192
x=192/8
x=24

angle 1 = 5x-12
=5(24)-12
=120-12
=108

angle 2= 3x =3*24=72

angle 1 = angle 9
angle 9= 108

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B2%7D%20%2B1%20%3D%20%5Csqrt%7B3%7D%20" id="TexFormula1" title=" \frac{x}{2

Tresset [83]

I hope this helps you

3 0

3 years ago

Simplify the expression:<br> (2p + 1)(2) =

Oxana [17]

Answer:

4p +2

Step-by-step explanation:

(2p + 1)(2)

Distribute

2p * 2 + 1*2

4p +2

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3 years ago

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Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

7 0

3 years ago