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2 years ago

6

X+4 what is the value of x

1 answer:

anzhelika [568]2 years ago

3 0

Answer:

Step-by-step explanation:

hello :

x+4=0

x= - 4

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First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

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3 years ago

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alexira [117]

Answer:

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7 0

2 years ago

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Lashaya decides to jog 8 miles. She is only able to jog 5/16 that distance. Write the distance

Novay_Z [31]

Answer:

The distance is $2.5\ miles$

Step-by-step explanation:

Let

x------> the distance

we know that

To find the distance, multiply the total distance of 8 miles by 5/16

so

$x=8(\frac{5}{16})=\frac{5}{2}=2.5\ miles$

6 0

2 years ago

At what position on the number line is the red dot located?

raketka [301]

Answer:

C. 0.5..................

3 0

2 years ago

A woman is randomly selected from the 18-24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally

dybincka [34]

Answer:

0.0274

Step-by-step explanation:

The mean is $\mu =114.8$ and the standard deviation is $\sigma =13.1.$

Calculate

$Z=\dfrac{X-\mu}{\sigma}$

for $X=140:$

$Z=\dfrac{140-114.8}{13.1}\approx 1.9237.$

If $X\sim N(114.8,\ 13.1),$ then $Z\sim N(0,1)$

and

$Pr(X>140)=Pr(Z>1.9237).$

Use table for normal distribution probabilities to get that

$Pr(Z>1.9237)=1-Pr(Z\le 1.9237)=1-0.9726=0.0274.$

7 0

3 years ago

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