<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4%20-%20%20%7Bx%7D%5E%7B2%7Ddx%20%7D%20" id="TexFormula1" title=" \sqrt{4

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Tamiku [17]

3 years ago

{2}dx } " alt=" \sqrt{4 - {x}^{2}dx } " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Ivan3 years ago

6 0

Answer:

do you know what x or dx =

Step-by-step explanation:

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Subtract. Write your answer in simplest form.<br><br> 9 1/2- 1 6/7

yaroslaw [1]

The answer is 9 1/2 - 1 6/7 = 4 1/8 hoped this helped

5 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

2 years ago

Factor 9y 2 - 12y 4.

vesna_86 [32]

Hope i understand it right

9y^2 -12y^4 = 3y^2 *(3 - 4y^2)

5 0

3 years ago

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determine whether the given ordered pair is a solution to the system of equations. 6x-2y=24 -3x+3y=24 and (10,18) yes, it is a s

Anika [276]

Answer:

I love algebra anyways

the ans is in the picture with the steps

(hope it helps can i plz have brainlist :D hehe)

Step-by-step explanation:

7 0

3 years ago

What is the range of possible sizes for side x?

faust18 [17]

0.5 < x < 16.5 given: Two sides of triangle: 8.0 units and 8.5 units
Measure of third side = x
According to the triangle's inequality,
Sum of any two sides > third side. (i)
Difference between the sides < third side. (ii)
If x is the third side, then
x < 8+8.5 [Using (i)]
i.e. x< 16.5
Also, x > 8.5-8 [Using (ii)]
i.e. x> 0.5
Hence, Range of possible sizes for side x would be 0.5 < x < 16.5.

6 0

2 years ago

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