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3 years ago

What is the area of the shaded region belo

Mathematics

1 answer:

ohaa [14]3 years ago

7 0

I can't tell. it's so far below that I can't see it.

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Please helppp<br> Find the amplitude of this function.

nordsb [41]

THe amplitude is half of the distance from the highest and lowest phase of the wave..

4-(-1)=5
0.5*5=2.5

amp=2.5 units

4 0

3 years ago

=<br> Drag up the correct symbol to make each statement true.

Dvinal [7]

Answer:

1. >

2.>

3.<

4. <

Step-by-step explanation:

I am doing it from top to bottom

I hope this helps! :)

4 0

2 years ago

X + 1 / 8 y = 2 0 x + 1 / 4 y = 4

UkoKoshka [18]

Answer:

Is that the question

Step-by-step explanation:

No seresly is that the question

4 0

3 years ago

Read 2 more answers

Using first principle find the derivative of y=3(2x+4)^2

vfiekz [6]

Answer:

fhgfcbhfftikjg_-+64-+--_-+6-&+

Step-by-step explanation:

5+-"_-&&76$&+655+65-+65&-5_

3 0

3 years ago

Solve x'=5t(sqrt(x)) x(0)=1

LekaFEV [45]

Answer:

$2\sqrt{x}=\frac{5t^2}{2}+2$

Step-by-step explanation:

Given: $\frac{\mathrm{d} x}{\mathrm{d} t}=5t\sqrt{x}\,,\, x(0)=1$

Solution:

A differential equation is said to be separable if it can be written separately as functions of two variables.

Given equation is separable.

We can write this equation as follows:

$\frac{dx}{\sqrt{x}}=5t\,dt$

On integrating both sides, we get

$\int \frac{dx}{\sqrt{x}}=\int 5t\,dt$

Formulae Used:

$\int \frac{1}{\sqrt{x}}=2\sqrt{x}\,\,,\,\,\int t\,dt=\frac{t^2}{2}$

So, we get solution as $2\sqrt{x}=\frac{5t^2}{2}+C$

Applying condition: x(0) = 1, we get $C=2$

Therefore, $2\sqrt{x}=\frac{5t^2}{2}+2$

8 0

2 years ago