(5×7) + 3 = 5 ×(2×7) true or false

Mathematics

2 answers:

svetlana [45]3 years ago

8 0

Answer:

false

Step-by-step explanation:

(5×7)+3

(35)+3

5×(2×7)

5×(14)

myrzilka [38]3 years ago

5 0

Step-by-step explanation:

35 + 3 = 5 x 14

38 = 70

False

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Can you guyz help me with this question.please write the answers step by step ☺️

sasho [114]

Hey! Ok so with this problem, you can figure out the pattern with subtraction.

If you convert the fractions to decimals you get 3.25 and 5.625.
5.625 - 3.25 = 2.375, which means he increased his run by 2.375 miles.
8 - 5.625 = 2.375 as well, which follows the pattern. So what you would need to do from here is add 2.375 miles for each day he jogged.

8 + 2.375 = 10. 375, the miles he jogged on thursday.
10.375 + 2.375 = 12. 75, the miles he jogged on friday.

So, when converted to fractions, your answer is D: Eric jogged 12 3/4 miles on Firday

7 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,

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

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