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Ilia_Sergeevich [38]

3 years ago

8

Determine polynomial expressions for the perimeter and the area of the figure shown below

1 answer:

kenny6666 [7]3 years ago

5 0

Perimeter: 4x + 2

The perimeter is equal to 2 times the length plus 2 times the width.

2(x + 1) + 2(x)

2x + 2 + 2x

4x + 2

Area: x^2 + x

The area is equal to the length times the width.

(x + 1)(x)

x^2 + x

Hope this helps! :)

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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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A bird flies from the bottom of a canyon that is 71 4/5 feet below sea level to a nest that is 875 7/10 feet above sea level. Wh

Rainbow [258]

The difference in elevation between the bottom of the canyon and the bird's nest is 947 1/2 feet

<h3>What is the difference in elevation between the bottom of the canyon and the bird's nest?</h3>

The given parameters are:

Nest = 71 4/5 feet above the seal level

Bottom of canyon = 875 7/10 below sea level

Below sea level means negative

So, we have:

Nest = 71 4/5 feet

Bottom of canyon = -875 7/10

The difference in elevation between the bottom of the canyon and the bird's nest is calculated as

Difference = |Nest - Bottom of canyon|

This gives

Difference = |71 4/5 - (-875 7/10)|

Evaluate the difference

Difference = |947 1/2|

Remove the absolute bracket

Difference = 947 1/2

Hence, the difference in elevation between the bottom of the canyon and the bird's nest is 947 1/2 feet

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1 year ago

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3 years ago

Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

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$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

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$x^3v=\displaystyle\int3x^2\,\mathrm dx$

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$v=1+\dfrac C{x^3}$

$y^3=1+\dfrac C{x^3}$

$\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}$

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3 years ago