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3 years ago

please help me, please I can’t fail this test!!❤️

Mathematics

2 answers:

barxatty [35]3 years ago

6 0

The answer I think is D because the middle is the median.

kodGreya [7K]3 years ago

3 0

Answer: i think A

Step-by-step explanation:

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3 years ago

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3 years ago

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3 years ago

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photoshop1234 [79]

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3 years ago

Read 2 more answers

In the midpoint rule for triple integrals we use a triple riemann sum to approximate a triple integral over a box b, where f(x,

Lana71 [14]

The sub-boxes will have dimensions $\frac{2-0}{2} \times \frac{2-0}{2} \times \frac{2-0}{2} =1\times1\times1=1 \ cubic \ units$

x sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $x= \frac{1}{2}$ and $x= \frac{3}{4}$
y sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $y= \frac{1}{2}$ and $y= \frac{3}{4}$
z sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $z= \frac{1}{2}$ and $z= \frac{3}{4}$ 

Let $f(x,y,z)=\cos{(xyz)}$

$\int\limits \int\limits \int\limits {f(x,y,z)} \, dV \approx f\left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{3}{4} \right)$
$+f\left( \frac{3}{4} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{3}{4} \right) \\ \\ \approx\cos{ \frac{1}{8} }+\cos{ \frac{3}{16} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{9}{32} }+\cos{ \frac{27}{64} } \\ \\ \approx0.9922+0.9825+0.9825+0.9607+0.9825+0.9607+0.9607 \\ +0.9123 \\ \\ \approx\bold{7.734}$

5 0

3 years ago

￼please help me, please I can’t fail this test!!❤️

please help me, please I can’t fail this test!!❤️