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1 year ago

!!50 POINTS!! need it pronto!!

Mathematics

1 answer:

valentinak56 [21]1 year ago

7 0

Answer:

I would like to help but it's hard tho sorry

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Jaylah has two packages of trail mix. The first package weighs 2.40 pounds and the second package weighs 3.25 pounds. She mixes

navik [9.2K]

She can make up to 22 bags because you have to add 2.40 and 3.25 together to get 5.65 then divide by .25 to get 22.6 she can make a total of 22 with .6 left over

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2 years ago

A regular hexagon has an apothem measuring 14 cm and an approximate perimeter of 96 cm.

Mariulka [41]

ANSWER

672cm²

EXPLANATION

The area of a regular hexagon is given by the formula,

$Area = \frac{1}{2} ap$

where a=14cm is the apothem and p=96cm is the perimeter of the regular hexagon.

We substitute the values into the formula,

$Area = \frac{1}{2} \times 14 \times 96 {cm}^{2}$

$Area =672 {cm}^{2}$

Therefore the approximate area of the hexagon is 672cm²

8 0

2 years ago

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A culinary club earns $1190 from a dinner service. They sold 34 adult meals and 51 student meals. an adult meal costs twice as m

Murrr4er [49]

59 gdgdksgfjjjkkkkjjjjjjjjjjjjjjjj

5 0

2 years ago

Read 2 more answers

Find the length of ac

maksim [4K]

Answer:

$\boxed{AC = 42.99}$

Step-by-step explanation:

Tan 34 = $\frac{opposite}{adjacent}$

Where opposite = 29, adjacent = AC

So,

0.674 = $\frac{29}{AC}$

AC = $\frac{29}{0.674}$

AC = 42.99

8 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

2 years ago