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KengaRu [80]
3 years ago
9

Para construir una estantería, un carpintero necesita lo siguiente:

Mathematics
1 answer:
yuradex [85]3 years ago
4 0
Nada más puede construir 5 estanterías completas.
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Find the distance d from (6.1, 5.9) to (-5.9, -4.1) .
photoshop1234 [79]
Distance formula:
√(x1-x2)^2+(y1-y2)^2

4 0
3 years ago
38. Which equation represents the data in<br> the table?
sergejj [24]

Answer:

C. y = 2x -4

Step-by-step explanation:

We can use the points (0, -4) and (2, 0).

y = mx + b (m is slope, b is y-intercept)

m = delta y / delta x

m = 4 / 2

m = 2

y = 2x + b

We can plug in (2, 0):

0 = 4 + b

b = -4

y = 2x - 4

6 0
3 years ago
Read 2 more answers
Can someone help please
Ymorist [56]

Answer:it is 7 and 6 that is your answer if that does not work it is 12 and 13 ok.

8 0
3 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
3 years ago
Solve the equation.<br> (x-5)(x+3)=0
LuckyWell [14K]
(x-5)(x+3)=0
x^2+3x-5x-15=0
(x^2+3x) (-5x-15)
x(x+3) -5(x+3)
x-5=0 x+3=0
x=5 OR -3
8 0
3 years ago
Read 2 more answers
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