Ask question

Ask question

All categories

2 years ago

10

Anita needs 140 cubes to make a model. The figure shows the number of cubes in a box How many boxes of cubes does Anita need to

buy so she has enough cubes for her model?

1 answer:

marishachu [46]2 years ago

3 0

Answer:

3 boxes

Step-by-step explanation:

First, we find how many cubes are in each box, then we find how many boxes you need for 140 cubes.

<u>To find the number of cubes in a box we need to find the volume of the figure</u>

the top is 2 wide and 4 long so there are 8 cubes on each layer. Sense it is 8 cubes tall with 8 cubes on each layer the model has 8×8=64 cubes in each box.

<u>Now we divide then round up to find how many full boxes she needs to get 140 cubes or more.</u>

If 1 box has 64 cubes and she needs 140 cubes 140/64=2.187. Now we round up sense he can't buy a fraction of a box and the answer is: She needs 3 boxes

You might be interested in

What is the size of angle x !!!

klemol [59]

Answer:

∠x = 67°

Step-by-step explanation:

∠x = 67° because they both are inscribed angles of the same arc

7 0

3 years ago

Read 2 more answers

Compute the probability of randomly selecting a three or club.

Pavlova-9 [17]

Find the total number of cards in a deck. Count all the # 3’s total cards/#3 cards
Count all the clubs total card/ clubs
Not sure if you are counting 3’s & clubs together. If you are, total cards/ both
So all you have to do is count & then divide to get the probability.

5 0

2 years ago

Solve the equation 58 - 10x ≤ 20 + 9x

drek231 [11]

Answer:

x≥2

Step-by-step explanation:

First, write out the equation as you have it:

$58-10x\leq20+9x$

Then, add $10x$ to both sides:

$58-10x+10x\leq20+9x+10x\\58\leq20+19x$

Next, subtract $20$ from both sides:

$58-20\leq20-20+19x\\38\leq19x$

Finally, divide both sides by $19$ :

$\frac{38}{19}\leq\frac{19x}{19}\\2\leq x$

or

$x\geq 2$

Therefore: the answer to this inequality/equation is: x≥2

6 0

3 years ago

Help I need to turn this in by 12:30

11Alexandr11 [23.1K]

B- -3

If x is two according to the coordinate points, that means the equation is now -8+3y=-17

Add 8 to each side- 3y=-9

And divide -9 by 3

This gets you y=-3

5 0

3 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago