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Anastasy [175]
2 years ago
6

Pls help i don’t get it , ill mark brainlist

Mathematics
2 answers:
Tcecarenko [31]2 years ago
8 0

Answer:

100 and 2

Step-by-step explanation:

professor190 [17]2 years ago
6 0

Answer:

100 and 2

Step-by-step explanation:

If we are trying to reduce the radical, we know that 100 times 2 is 200.Therfore,Those are two possible numbers to reduce the radical.

                                             Hope this is right !

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16
4vir4ik [10]

Answer:

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

9 -16 (x^2 + y^2) = 0 \\ \\  16 (x^2 + y^2)  = 9 \\ \\  x^2+y^2 = \dfrac{9}{16}

x^2+y^2 = (\dfrac{3}{4})^2

where:

0 ≤ r ≤ \dfrac{3}{4} and 0 ≤ θ ≤ 2π

∴

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0}  \ \ \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0}  ( 9r^2-16r^4})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

4 0
3 years ago
PLEASE HELP I'LL MARK YOU BRAINLIEST IF YOU'RE CORRECT I don't understand this​
mario62 [17]
The y-intercept of the function is at y = -3 so the options that don’t include it are off the table. The only possible answer is option C. Option A doesn’t work because plugging in positive values would give negative numbers within the square root which isn’t defined in the real numbers.
8 0
2 years ago
Read 2 more answers
In AABC, AC - BC. The length of AC is three times the length of AB. Find
Arlecino [84]

Answer:

i don't undertsand this question

5 0
3 years ago
HELP PLEASE<br> What is the following product?
Naya [18.7K]
The answer would be: 

6 0
3 years ago
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Let g (x) = 3x+2 and f (x) = x-2 / 3 find the value. g (f(2))
Romashka [77]

Answer: OPTION C.

Step-by-step explanation:

In order to solve the given exercise, you can follow these steps:

1. Given the following function f(x):

f(x)=\frac{x-2}{3}

You must substitute x=2 into the function f(x). Then:

f(2)=\frac{(2)-2}{3}

2. Evaluating, you get:

f(2)=\frac{0}{3}\\\\f(2)=0

3. Now, the next step is to substitute f(2)  into the function g(x):

g (x) = 3x+2\\\\g (f(2)) = 3(0)+2

4. Finally, evaluating, you get the following result:

g (f(2)) = 0+2\\\\g (f(2)) =2

You can identify that it matches with the Option C.

3 0
3 years ago
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