Dark : Light
3 : 12
(÷3) (÷3)
1 : 4

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Answer : The ratio is 1:4
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6 0

3 years ago

Read 2 more answers

A circle has a circumference of 50.2450.2450, point, 24 units.<br> What is the radius of the circle?

beks73 [17]

Answer:

r = 8 units

Step-by-step explanation:

The circumference is calculated using $C = 2\pi r$ . Substitute C = 50.24 units. Solve for the radius.

$C = 2 \pi r\\50.24 = 2 \pi r\\25.12 = \pi r\\8 = r$

7 0

3 years ago

The perimeter of a triangle is 49ft. One side of the triangle is 17ft shorter than the second side. The third side is 6ft shorte

Phoenix [80]

Answer:

7,24,18

Step-by-step explanation:

perimeter of a triangle = a₁ + a₂ + a₃

A/Q

let, 2nd side = x

then, 1st side =x-17

3rd side = x-6

perimeter= 49

49 = (x-17)+ x + (x-6)

49 = x-17+x+x-6

49 = 3x -23

3x = 49+23

3x = 72

x = 72/3

x= 24

1st side = (x-17) = 24-17=7

1st side= 7

2nd side = x = 24

3rd side = x-6

=24-6

=18

3rd side = 18

on calculating perimeter = 7+24+18=49.

Hope it helps:)

8 0

3 years ago

Given vectors u = ⟨–3, 2⟩ and v = ⟨2, 1⟩, what is the measure of the angle between the vectors?

hjlf

The measure of the angle between the vectors

$$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }$ .

What is the measure of the angle between the vectors?

Given:

$\mathrm{u}=\langle -3,2\rangle$ and $v=\langle 2,1\rangle$

Computing the angle between the vectors, we get

$\quad \cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$

To estimate the lengths of the vectors, we get

Computing the Euclidean Length of a vector,

$\left|\left(x_{1}, \ldots, x_{n}\right)\right|=\sqrt{\sum_{i=1}^{n}\left|x_{i}\right|^{2}}$

Let, $$\mathrm{u} &=\langle -3,2\rangle \\$ and $$\mathrm{v} &=\langle 2,1\rangle \\$

If $$\mathrm{u} &=\langle -3,2\rangle \\$

$$|u| &=\sqrt{-3^{2}+(2)^{2}} \\$

$$&=\sqrt{5}i \\$ and

$$\mathrm{v} &=\langle 2,1\rangle \\$

$$|v| &=\sqrt{2^{2}+(1)^{2}} \\$

$$&=\sqrt{5}$

Finally, the angle is given by:

Computing the angle between the vectors, we get

$$\cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$

$$&\cos (\Phi)=-\sqrt{13 } i/ \sqrt{5 } \\$

simplifying the above equation, we get

$$&\Phi=\arccos (\cos (\Phi))$

$$=\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }$

Therefore, the measure of the angle between the vectors

$$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }$ .

To learn more about vectors refer to:

brainly.com/question/19708440

#SPJ4

8 0

2 years ago