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Greeley [361]
3 years ago
5

The pictogram shows information about the numbers of phones sold in a shop over a 5-month period.

Mathematics
1 answer:
OLEGan [10]3 years ago
7 0

Answer:

a) 5(6) = 30

b) 3.5(6) = 21

c) 153

Step-by-step explanation:

c) 30 + 39 + 21 + 36 + 27 = 153

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Amiraneli [1.4K]

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cos(\sqrt{\frac{3}{2} }) or 0.339

Step-by-step explanation:

6 0
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Find the surface area of regular prism with height 6 cm, if the base of the prism is a Regular quadrilateral with side 4 cm.
marshall27 [118]
A "regular quadrilateral" is a square, so the length and width are both 4 cm. The surface area of a rectangular prism is given by
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6 0
2 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
2 years ago
Help me with Thai math it’s worth 22 points PLZZ answer ALL of them
joja [24]

Answer:

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