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3 years ago

Help please! No file attachments please, only write your answers in response to question.

Mathematics

1 answer:

brilliants [131]3 years ago

4 0

This is the correct answer for the problem
AFK,CGJ,DEL,BHM

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HELP ME PLZZ I NEED HELP WITH THIS!!

Aleonysh [2.5K]

Answer:

its the second one

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The cost of petrol rises by 2 cents a liter. Last week a man bought 20 liters at the old price. This week he bought 10 liters at

In-s [12.5K]

Answer: the old price for 1 liter is $0.3

Step-by-step explanation:

Let x represent the old price for 1 liter of petrol.

Last week a man bought 20 liters at the old price. This means that the total amount that the man paid for 20 liters is 20x.

The cost of petrol rises by 2 cents a liter. This means that the new price per liter is x + 2/100 = x + 0.02

This week he bought 10 liters at the new price. This means that the total amount that the man paid for 10 liters is

10(x + 0.02)

= 10x + 0.2

Altogether, the petrol costs $9.20. This means that

20x + 10x + 0.2 = 9.2

30x + 0.2 = 9.2

30x = 9.2 - 0.2

30x = 9

x = 9/30

x = 0.3

3 0

3 years ago

Help please...................

dlinn [17]

Answer:

2.8 inches each month

Step-by-step explanation:

14 divided by 5

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3 years ago

Write down, in details the outcomes of Ordinary Differential Equations and Special<br> Functions

Anastaziya [24]

Answer:

an equation that involves some ordinary derivatives (as opposed to partial derivatives

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3 years ago

Help please and thank you

Dominik [7]

Answer:

Step-by-step explanation:

We assume the rotation R is <em>counterclockwise</em> 60°.

The exponent on R is the number of times it is applied. That is, R² = R(R(figure)). So, the composition is equivalent to R^(2-4) = R^-2.

When the exponent of R is negative, it is essentially the inverse function. That is, applying the function R to the result will give the figure you started with. Equivalently, it is rotation in the other direction.

$(\mathcal{R}^2\circ\mathcal{R}^{-4})(\text{figure})=\mathcal{R}^{-2}(\text{figure})=\text{figure rotated $120^{\circ}$ CW}$

The point 120° clockwise from B is D.

The desired image point is D.

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3 years ago