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Free_Kalibri [48]
3 years ago
9

Help help help help help pls

Mathematics
1 answer:
nekit [7.7K]3 years ago
3 0
The range is 20 darlin
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250 gram as a percentage of 1 kg ​
kodGreya [7K]

Hey there!

Answer: 25%

\rightarrow \:  (\sf{ \frac{250}{1000} ) \times 100}

\rightarrow \: ( \sf{0.25) \times 100}

\rightarrow \:  \sf{ = 25\%}

5 0
2 years ago
Read 2 more answers
What property is being illustrated below?<br> 2(7+3)=2(7)+2(3)
AleksAgata [21]

Answer: Associative property

Step-by-step explanation:

8 0
2 years ago
Evaluate m – n if m = –16 and n = 28
natima [27]

Answer:-44

Step-by-step explanation:

-16-(28)=-44

4 0
2 years ago
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Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
Write the common rule to describe the translation using coordinate notation.
UNO [17]

The common rule is (x - 1, y + 3) can be used to describe the translation.

Step-by-step explanation:

Step 1:

The point J (-3, -4) becomes J^{1} (-4, -1).

In order to write the rule for translation from J to J^{1}, we subtract the x coordinate of J from J^{1} and subtract the y coordinate of J from J^{1}.

The x coordinate = -4-(-3) = -4+3 = -1.

The y coordinate = -1-(-4) = -1+4 = 3.

Step 2:

So from the calculations, we get that the x coordinate is subtracted by 1 i.e. x-1 and the y coordinate is increased by 3 i.e. y+3.

So the common rule is (x - 1, y + 3).

4 0
3 years ago
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