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4 years ago

11

Markers are sold in boxes, packs, or as single markets. Each box has 10 packs. Each pack has 10 markers. Draw a picture to show

two ways to buy 276 markers

2 answers:

Neko [114]4 years ago

7 0

Answer:

Two possible ways are:

2 boxes, 7 packs and 6 single markers

2 boxes 6 packs and 16 single markers.

Step-by-step explanation:

Consider the provided information.

We have 276 markers. Each box has 10 packs, and each pack has 10 markers

We need to put 276 markers in boxes packs or single.

One possible way is:

Each box has 10 packs, and each pack has 10 markers; it means each box holds 10(10) = 100 markers.

Two boxes contains 2(100) = 200 markers.

Now the markers left are:

276-200 = 76 markers.

Each pack has 10 markers, therefore 7 packs would be 7(10) = 70 markers. This would leave

76-70 = 6 markers.

So, the one possible way is 2 boxes 7 packs and 6 single markers.

Way 2:

If we choose to use 2 boxes and 6 packs, this would be

2(100) + 6(10) = 200 + 60 = 260 markers,

This would leave

276-260 = 16 markers.

The another possible way is 2 boxes 6 packs and 16 single markers.

The required figure is shown below:

Lapatulllka [165]4 years ago

5 0

You could buy 276 single markers, you can buy 28 packs of markers, or 3 boxes of marker because 1pack = 10 markers, 10 markers multiplied by 10 packs equals 100. So 1 box would have 100 markers in it and if you took 2 boxes you would be missing 76 markers, so you're better off getting 3 boxes and having 300 markers than being less by 76.

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3 years ago

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Let |u| = 10 at an angle of 45° and |v| = 13 at an angle of 150°, and w = u + v. What is the magnitude and direction angle of w?

mixer [17]

Given:

$|u|=10$ at an angle of 45°.

$|v|=13$ at an angle of 150°.

$w=u+v$

To find:

Magnitude and direction angle of w.

Solution:

We have,

$w=u+v$

$\theta = 45^\circ, \phi = 150^\circ$

Now,

$u_x=|u|\cos \theta=10\cos 45^\circ=7.071$

$u_y=|u|\sin \theta=u_y=10\sin 45^\circ=7.071$

$v_x=|v|\cos \phi=13\cos 150^\circ=-11.25833$

$v_y=|v|\sin \phi=u_y=13\sin 150^\circ=6.5$

Using these information, we get

$R_x=u_x+v_x=7.071-11.25833=-4.18733$

$R_y=u_y+v_y=7.071+6.5=13.571$

$\text{Direction angle}=\tan^{-1}(\dfrac{R_y}{R_x})$

$\text{Direction angle}=\tan^{-1}(\dfrac{13.571}{-4.18733})$

$\text{Direction angle}=-72.8239$

$\text{Direction angle}=107.1476$

$\text{Direction angle}\approx 107.1$

Now,

$|w|=\sqrt{(|u|\cos \theta +|v|\cos \phi)^2+(|u|\sin \theta +|v|\sin \phi)^2}$

$|w|=\sqrt{(10\cos(45)+13\cos 150)^2+(10\sin(45)+13\sin 150)^2}$

$|w|=\sqrt{(10(\dfrac{1}{\sqrt{2}})+13(-\dfrac{\sqrt{3}}{2}))^2+(10(\dfrac{1}{\sqrt{2}})+13(\dfrac{1}{2}))^2}$

$|w|=14.20236$

$|w|\approx 14.20236$

Therefore, the correct option is D.

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