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3 years ago

How to get smarter in math?

Mathematics

2 answers:

anastassius [24]3 years ago

8 0

Practice, practice & practice.

frutty [35]3 years ago

3 0

You could arrange for a tutor.

You could ask your guardian to help explain.

You could ask your teacher.

Practice and study what you are having difficulty with.

Remember: PRACTICE MAKES PERFECT!!!

(not always but you will get better)

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Find Value Of X. Help haahahhnak

Sergeu [11.5K]

Don’t click the link.

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3 years ago

Solve by addition -8x+2y=14 -4x+y=4

Mumz [18]

Answer:Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.

No solution

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the weight, in Newtons, of a person on Earth whose mass is 90 kilograms? Gravity near the Earth's surface is 9.8 m/s2.

valentinak56 [21]

Answer:

W = 882 N

Step-by-step explanation:

Given that,

The weight of a person, W = 90 kg

The value of acceleration due to gravity, g = 9.8 m/s²

We need to find the weight of the person on Earth. The formula for the weight of an object is given by :

W = mg

Substitute all the values,

W = 90 kg × 9.8 m/s²

W = 882 N

So, the weight of the person is equal to 882 N.

8 0

3 years ago

What percent of 30 is 33

Llana [10]

Hi!

To solve this we can set up a proportion and cross multiply.

$\frac{33}{30} \frac{x}{100}$

33 x 100 = 3300
3300/30 = 110

The answer is 110%

Hope this helps! :)

3 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

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

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago