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stich3 [128]
3 years ago
7

If the measure of an angle 38 , find the measure of its complement

Mathematics
1 answer:
Darya [45]3 years ago
5 0
If you’re finding the measure of its complement, the sum of both angles will always be 90*

x+38 = 90
-38 | -38

x = 52

the measure of its complement is 52*
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Which best describes the relationship between the successive terms in the sequence shown?
Alex Ar [27]

Given series is 2.4,-4.8,9.6,-19.2

To find whether it has common difference or common ratio let us find few differences and few ratios of consecutive terms.

Common difference of first 2 terms = 2nd term - first term = -4.8-2.4 = -7.2

Common difference of 2nd and 3rd terms = 3rd term - 2nd term = 9.6-(-4.8) = 14.4

Since those common differences are not equal the given series does not have common difference at all.

To check if it has common ratio or not let us find few ratios of consecutive terms.

Common ratio of first 2 terms = \frac{2nd term}{first term} = \frac{-4.8}{2.4} = -2

Common ratio of 2nd and 3rd terms = \frac{3rd term}{2nd term} = \frac{9.6}{-4.8} = -2.0

So, the given series has common ratio as -2.0

7 0
3 years ago
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Point A is located at -5/8 and point B is located at -2/8. What is the distance between points A and B
UNO [17]
I think those are coordinates of points, (-5,8) and (-2,8)
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square root 9=3
3 0
3 years ago
Calculate the minimum value of the quadratic f(X)=x^2-6x+10<br>​
tangare [24]
The answer is (-3,-1)
8 0
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A popular online music store allowed shoppers to listen to an entire album before buying it. Of the people who listened to the w
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4 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
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