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Luden [163]
3 years ago
11

Pleeeeeese helppp!!!!!!!!!!!

Mathematics
1 answer:
Svetlanka [38]3 years ago
7 0
I think 17.2 sorry if im wrong

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Suppose y varies directly with x and y=20 when x = 10. What is the value of y<br> when x = 30?
S_A_V [24]

Answer:

y = 60

Step-by-step explanation:

Given that y varies directly with x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 20 when x = 10, then

20 = 10k ( divide both sides by 10 )

2 = k

y = 2x ← equation of variation

When x = 30, then

y = 2 × 30 = 60

7 0
3 years ago
Please help me with this
docker41 [41]

Answer: y=12.287

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
(-1/4)^8 * (8/11)^6 * (1/2)^-8
marissa [1.9K]

Answer:

1024/1771561

Step-by-step explanation:

Using exponent rules, we can multiply all of them together after applying the exponent to get the final answer.

8 0
4 years ago
Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified, exact value in terms o
Digiron [165]

Check the picture below.

\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}

notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.

7 0
3 years ago
Amy purchased a brand-new mountain bike, which she's riding through
Neko [114]

Answer:

The Acceleration of bike after 3 minutes of ride is  14 meter² per minutes

Step-by-step explanation:

Given as :

The displacement of bike is the function of time

i.e s(t) = t³ - 2 t² + 3  meters

The Time of bike acceleration = t = 3 minutes

Let The Acceleration of bike = a meter² per minutes

Now, According to question

∵ <u>Acceleration is define as the rate of change of velocity</u>

i.e acceleration = \dfrac{\textrm velocity}{\textrm time}

or, a =  \dfrac{\textrm v}{\textrm t}

And

<u>Velocity is define as rate of change of displacement</u>

i.e velocity =  \dfrac{\textrm displacement}{\textrm time}

Or, v =  \dfrac{\textrm s}{\textrm t}

Since here displacement is function of time

So , v = \frac{\partial s(t)}{\partial t}

Or, v =   \frac{\partial (t³ - 2 t² + 3)}{\partial t}

Or, v = 3 t² - 4 t + 3

So, velocity is the function of time = v = 3 t² - 4 t + 3   meter per minutes

Now, Again

Acceleration = a = \frac{\partial v}{\partial t}

Or, a = \frac{\partial (3 t² - 4 t + 3)}{\partial t}

Or, a = 6 t - 4

∴  Acceleration is the function of time = a = 6 t - 4

Now, Acceleration of bike after 3 minutes

So, at t = 3 min

i.e a = 6 t - 4

Or, a = 6 × 3 - 4

Or, a = 18 - 4

∴  a = 14 meter² per minutes

Hence, The Acceleration of bike after 3 minutes of ride is  14 meter² per minutes . Answer

5 0
3 years ago
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