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Brums [2.3K]
3 years ago
10

Show two ways that six people can share an eight segment

Mathematics
1 answer:
melamori03 [73]3 years ago
4 0
Each person coulg get one and you would have 2 left over or you can divide the 2 into 6 even peices
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Write the sentence as an equation.<br><br><br> 328 is the quotient of 36 and b<br><br> FAST
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328 = \dfrac{36}{b}
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G(-4, -8) and H(-15, -3). What is the slope of GH?
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Answer:

G(x1,y1)= (-4,-8)

H(x2,y2)=(-15,-3)

Slope = (y2-y1)/(x2-x1)

=(-3-(-8))/ (-15-(-4))

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=5/-11

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Which expression is equivalent to 20?
DaniilM [7]

Answer:

C. 4 X (3 + 8 – 6)

Step-by-step explanation:

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8 0
3 years ago
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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
Subtract.
jeka94
(x+1)-(-2x-5)
=x+1+2x+5
=3x+6
=3(x+2)
7 0
3 years ago
Read 2 more answers
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