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3 years ago

9

Can you guys pls help me out

2 answers:

Ksju [112]3 years ago

6 0

Answer:

elderly

Step-by-step explanation:

its asking who needs it the most

Because the elderly are old and they are gonna die without life insurance cause they wont get there meds

Y_Kistochka [10]3 years ago

5 0

The answer would be elderly

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Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1

Charra [1.4K]

Separating variables, we have

$\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx$

Integrate both sides.

$\displaystyle \int y\,dy = \int x\,dx$

$\dfrac12 y^2 = \dfrac12 x^2 + C$

Given that $y(0)=-1$ , we find

$\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12$

Then the particular solution is

$\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12$

$y^2 = x^2 + 1$

$y = \pm\sqrt{x^2 + 1}$

and because $y(0)=-1$ , we take the negative solution to accommodate this initial value.

$\boxed{y(x) = -\sqrt{x^2+1}}$

7 0

2 years ago

PLEASE HELP I GIVE THANKS

7nadin3 [17]

The base of the tent is 72 across the whole way and the hypotenuse is 45
so
to find the length of the stick call it x
half 72 to reduce it to the right hand side triangle
45²-36²=x²
x²=729
x=27 inches<span />

5 0

3 years ago

Read 2 more answers

Expand the expression (r-4)(8)

Gnom [1K]

R - 4 x 8 is the expanded expression

7 0

3 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

Which of the following is not a requirement of a standard form equation Ax+By=C? a)A ≥ 0. b)B ≥ 0. c)A and B are not both 0. d)A

Ghella [55]

<h2>Answer:</h2>

The following which is not a requirement of a standard form of equation Ax+By=C is:

b) B ≥ 0

<h2>Step-by-step explanation:</h2>

We know that the standard equation of a line is given by:

$Ax+By=C$

where A,B and C are integers and A is taken to be a non-negative integer i.e. (A≥0) also the greatest common factor of A,B and C is: 1.

Also A and B can't be both zero.

Hence, the correct option is:

Option: b

7 0

3 years ago

Read 2 more answers