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3 years ago

10

Brian has reduced his cholesterol level by 8% after his last check up. If his original level was 200, what is his cholesterol le

vel now?
A.
198
B. 184
c. 216
D. 16

1 answer:

s2008m [1.1K]3 years ago

6 0

Answer:

B. 184

Step-by-step explanation:

A proportion is an easy way to solve this:

$\frac{100\%}{200} = \frac{92\%}{x} \\ 100x = 92(200) \\ x = 92(2) \\ x = 184$

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Answer:

$4

Step-by-step explanation:

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2 years ago

When cut into 10 equal slices each slice is 1/10 of the pizza Express how much of the pizza 7 of these slices would be in decima

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2 years ago

the volume of a cube is increasing at a constant rate of 824 cubic centimeters per minute. At the instant when the volume of the

Maurinko [17]

Answer:

DA/dt = 75.27 cm²

Step-by-step explanation:

Cube Volume = V(c) = 683 cm³

DV(c) /dt = 824 cm³

V(c,x) = x³

Then

DV(c,x)/ dt = 3x² Dx/dt

( DV(c,x)/ dt )/ 3x² = Dx/dt (1)

Now as V(c,x) = x³ when V(c,x) = 683 cm³ x = ∛683

x = 8.806 ( from excel)

And by subtitution of this value in equation (1)

Dx/dt = ( DV(c,x)/ dt )/ 3x² ⇒ Dx/dt = 824 / 3*x²

Dx/dt = 824 /3*77.55

Dx/dt = 824/232,64

Dx/dt = 3,542

Then we got Dx/dt where x is cube edge. The area of the face is x² then

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2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

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2 years ago

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Ksju [112]

Answer:

BAngle AOB is half as big as angle COB.

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