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2 years ago

Please help right now

Mathematics

2 answers:

Lelu [443]2 years ago

4 0

Answer:

5 and 6

Step-by-step explanation:

Alex2 years ago

3 0

Step-by-step explanation:

$\sqrt{26} is \: between 5.0and5.1$

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every year, Hannah sells blueberry and strawberry jam at the local farmers market. Hannah charges four dollars Per jar of bluebe

bazaltina [42]

Answer: part A- $4b+6s=c$

part B- $4[3]+6[5]=42$

Step-by-step explanation:

3 0

3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

is vector v with an initial point of (-5,22) and a terminal point of (20,60) equal to vector u with an intintal point of (50,120

marta [7]

Answer:

Yes, vectors u and v are equal.

Step-by-step explanation:

We need to check whether vectors u and v are equal or not.

If the initial point is $(x_1,y_1)$ and terminal point is $(x_2,y_2)$ , then the vector is

$Vector=(x_2-x_1)i+(y_2-y_1)j$

Vector v with an initial point of (-5,22) and a terminal point of (20,60).

$\overrightarrow v=(20-(-5))i+(60-22)j$

$\overrightarrow v=25i+38j$ ..... (1)

Vector u with an initial point of (50,120) and a terminal point of (75,158).

$\overrightarrow u=(75-50)i+(158-120)j$

$\overrightarrow u=25i+38j$ .... (2)

From (1) and (2) we get

$\overrightarrow u=\overrightarrow v$

Therefore, vectors u and v are equal.

5 0

3 years ago

Write the standard equation for each circle:<br> Center at (-1,6) with radius 1/3

Sunny_sXe [5.5K]

The answer is:
(X-(-1))^2 + (Y-(6))^2 = (1/3)^2

this can be simplified into the following
(X+1)^2 + (Y-6)^2 = 1/9

7 0

3 years ago

The intercepts of the equation x^2 + 4y^2 = 16 are ___<br> PLEASE HELP

Eva8 [605]

Yintercepts are where x=0
set to zero to solve

0^2+4y^2=16
4y^2=16
divide 4
y^2=4
sqrt both sides
y=+/-2

xintercept is where y=0

x^2+4(0)^2=16
x^2+0=16
x^2=16
sdqrtboth sides
x=+/-4

yintercepts are (0,2) and (0,-2)
xintercepts are (4,0) and (-4,0)

3 0

3 years ago