All categories

2 years ago

Beno

Mathematics

1 answer:

anastassius [24]2 years ago

5 0

Answer:

$ 14

Step-by-step explanation:

12 - 5 = 7 = half weekly allowance

You might be interested in

Please PLEASE HELP><br> What is the total bill you would leave for a $22 haircut and a 15% tip?

geniusboy [140]

Answer:

$25.30

Step-by-step explanation:

15% of 22 = 3.3

22 + 3.3 = 25.30

7 0

2 years ago

3 more than the product of 3 and y

Fudgin [204]

Answer:

3y+3

Step-by-step explanation:

pls give brainly

4 0

2 years ago

Brad earns $12.75/hr and works 40 hours a week. What is his gross annual income?

Mrac [35]

$12.75/hr x 40hrs = $510

6 0

3 years ago

Please help me with this please and thank you

Alina [70]

Step-by-step explanation:

y = ax^n + bx^(n-1) ... + z

(0, 1) tells us that the constant term z = 1.

because x = 0 eliminates all other terms. and as the sum is 1, that means z = 1.

for the other pairs I strongly suspect

(-1, 3) : 2×(-1)² + 1 = 2×1 + 1 = 3

(-2, 9) : 2×(-2)² + 1 = 2×4 + 1 = 9

so, the fitting equation is

y = 2x² + 1

3 0

2 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

2 years ago