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2 years ago

Can u help me ASAP!!!

Mathematics

1 answer:

Rom4ik [11]2 years ago

8 0

Can be reduced to 1/3

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Micheal has a rectangular shed of an area of 24 square feet. The shed is 5 feet longer than it’s width. Create an equation that

BaLLatris [955]

Answer:24= x+5

Step-by-step explanation:

7 0

2 years ago

Cot^2x/cscx-1=1+sinx/sinx

KATRIN_1 [288]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------\\\\
\cfrac{cot^2(x)}{csc(x)-1}=\cfrac{1+sin(x)}{sin(x)}\impliedby \textit{let's do the left-hand-side}$

$\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)}
\\\\\\
\cfrac{cos^2(x)}{sin(x)}\cdot \cfrac{1}{1-sin(x)}\implies \cfrac{cos^2(x)}{sin(x)[1-sin(x)]}$

$\bf \cfrac{1-sin^2(x)}{sin(x)[1-sin(x)]}\implies \cfrac{1^2-sin^2(x)}{sin(x)[1-sin(x)]}
\\\\\\
\cfrac{\underline{[1-sin(x)]}~[1+sin(x)]}{sin(x)\underline{[1-sin(x)]}}\implies \cfrac{1+sin(x)}{sin(x)}$

5 0

2 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

2 years ago

Tell me the answers please!

il63 [147K]

18 is D and 19 is C you're welcome

4 0

3 years ago

It is desired to draw blood 3 cm up a capillary tube. Assuming a contact angle of 0 degrees and a blood density of 1060 kg/m3, w

JulijaS [17]

Answer:

If we assume a temperature of 20ºc and the blood interfacial surface tension is similar to water interfacial surface tension, the diameter of the capillary tube should be 0.933mm.

Step-by-step explanation:

The Jurin law describes the height a fluid can reach in a capillary tube. This law can be written as:

$H=\displaystyle\frac{2\gamma cos(\theta)}{\rho gr}$

where γ is the interfacial surface tension, θ is the contact angle with the fluid, ρ is the fluid density, g is the gravity acceleration and r is the tube radius.

If we assume that the interfacial surface tension of blood and water are almost the same, γ=0,0728 N/m at 20ºc. Therefore the diameter of the tube will be:

$D=2r=\displaystyle\frac{4\gamma cos(\theta)}{\rho gH}=0.933\cdot 10^{-3}m$

4 0

2 years ago