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AveGali [126]
3 years ago
13

The area of triangle PQR is 231 cm2 , and PQ = 21 cm. Find the altitude SR. Help me solve this

Mathematics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

SR = 22 cm

Step-by-step explanation:

area = bh/2

(21 cm)(h)/2 = 231 cm^2

21h = 462 cm

h = 22 cm

SR = 22 cm

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Answer:

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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x
Serggg [28]

I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

so we first compute the curl:

\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

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\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

Then the line integral is equal in value to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}

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Answer:

The value of x is: x > -2

Step-by-step explanation:

The interval notation if you need it is (-2, ♾)

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3 years ago
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In mathematics, the term "center of dilation" refers to a constant point on a surface from which all other points are either enlarged or compressed. The center of dilation and the scale factor comprise the two properties of a dilation.

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3 years ago
Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:
tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

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Sort the dataset in ascending order

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<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

Q1 = \frac{N +1}{4}th

So, we have:

Q1 = \frac{13 +1}{4}th

Q1 = \frac{14}{4}th

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This is the average of the 3rd and the 4th element

Q1 = \frac{1}{2} \times (84 + 91)

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<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

Q3 =  3 \times \frac{N +1}{4}th

So, we have:

Q3 =  3 \times \frac{13 +1}{4}th

Q3 =  3 \times 3.5th

Q3 =  10.5th

This is the average of the 10th and the 11th element.

Q_3 =\frac12 \times (98 + 101)

Q_3 =99.5

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

IQR = Q_3 -Q_1

IQR = 99.5 - 87.5

IQR = 12

Also, we have:

IQR(1.5) = 12 \times 1.5

IQR(1.5) = 18

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

Lower = Q_1 - IQR(1.5)

Lower = 87.5- 18

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Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

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