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3 years ago

- 8x + 9y + 92 = -2 8x + y + 4z = 3 - 8x + 8y + 9z = 7

Mathematics

1 answer:

Lady_Fox [76]3 years ago

8 0

Answer:

- 8x + 8y + 9z = 7 i think that is the answer of this question

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Determine whether the number is a solution of the equation.

aev [14]

Answer:
Yes. -4 is the solution to the equation.
Explanation:
2 + 4(-4) = - 14
2 + (-16) = - 14
2 - 16 = - 14
- 14 = - 14

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3 years ago

What is the value of x?

Sever21 [200]

Answer:

x = 30 degrees

6 0

4 years ago

(52+2)-3x -6<br><br> help me with thanks

Yuri [45]

Answer:

48 - 3X

Step-by-step explanation:

( 52+2) - 3x - 6

54 - 3x - 6 So first we deal with the numbers in brackets and that is 52 + 2 giving us 54.

54 - 6 - 3x Then you simplify the expression that is collecting like terms so then we subtract 6 from 54

48 - 3x This is the final expression after simplifying

HOPE THIS HELPED

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3 years ago

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

Logan took out $23,400 in student loans to attend college at a compound interest rate of 5%. He deferred payments for two years.

blagie [28]

Answer:

$25,740

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 5%/100 = 0.05 per year,

then, solving our equation

I = 23400 × 0.05 × 2 = 2340

I = $ 2,340.00

The simple interest accumulated

on a principal of $ 23,400.00

at a rate of 5% per year

for 2 years is $ 2,340.00.

8 0

3 years ago