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stepan [7]
4 years ago
8

The glacier moves 5 centimeters in 8 day. How many days does it take the glacier to move 1 centimeter?

Mathematics
1 answer:
pantera1 [17]4 years ago
5 0
It will take 1.6 days to move
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Find the values of p and q in the figure.​
vivado [14]

Answer:

25 and 20

Step-by-step explanation:

p²=24²+7² ⇒ p²= 625 ⇒ p=25

q²=p²-15² ⇒ q²=625 - 225 ⇒ q²=400 ⇒ q= 20

6 0
4 years ago
a snail going full speed was taking 1/8 of a minute to move 1/7 of a centimeter. at this rate, how long would it take the snail
Lana71 [14]
It will take 7/8 of a minute because you just multiply 1/8 by 7
4 0
4 years ago
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In a triangle the first angle is greater than the second angle by 18 degree and less than the third angle by 10 grade. Express a
serg [7]

Answer:

The 3 angles are 62 2/3, 44 2/.3 and 72 2/3 degrees.

Step-by-step explanation:

let x be the first angle then the  the second angle is  x - 18 and the third angle = x + 10

The 3 angles add up to 180 so:

x + x - 18 + x + 10 = 180

3x = 180 + 18 - 10 = 188

x = 188/3 = 62 2/3 degrees.

x - 18 = 44 2/3

x + 10 = 72 2/3.

8 0
3 years ago
Given the equation 3x − 4y = 8, which equation below would cause a consistent-dependent system? 3x + 4y = −8 6x − 8y = 12 9x − 1
Andru [333]

3x - 4y = 8

y = mx + b ; m is the slope ; b is the y-intercept

-4y = -3x + 8
y = -3x/-4 + 8/-4
<span>y = 3/4 x - 2</span>
5 0
3 years ago
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For vectors u= (3,4) and v= (1,3) find CompuV and the angle between u and v.
vampirchik [111]

Answer:

The angle between the given vectors u and v is \theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]

Step-by-step explanation:

Given vectors are \overrightarrow{u}=(3,4) and \overrightarrow{v}=(1,3)

Now compute the dot product of u and v:

\overrightarrow{u}.\overrightarrow{v}=(3,4).(1,3)

  =(3)(1)+(4)(3)

  =3+12

 =15

Now find the magnitude of u and v:

|\overrightarrow{u}|=\sqrt{3^2+4^2}

=\sqrt{9+16}

=\sqrt{25}

=5

|\overrightarrow{u}|=5

|\overrightarrow{v}|=\sqrt{1^2+3^2}

=\sqrt{1+9}

=\sqrt{10}

|\overrightarrow{v}|=\sqrt{10}

To find the angle between the given vectors

\overrightarrow{u}.\overrightarrow{v}=|\overrightarrow{u}|\overrightarrow{v}|cos\theta

\theta=cos^{-1}\left[\frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{u}|\overrightarrow{v}|}\right]

=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]

=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]

\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]

Therefore the angle between the vectors u and v is

\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]

3 0
3 years ago
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