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2 years ago

Help me I'm stuck perhaps a little while

Mathematics

2 answers:

Finger [1]2 years ago

4 0

The answer is 6 1/3 u have to simplify

lys-0071 [83]2 years ago

3 0

Shouldn't it be 5 4/3?

You might be interested in

Find a formula for the inverse of the function. f(x)=1-2/x^3.

Elena-2011 [213]

f(x) = 1 - ²/ₓ₃
y = 1 - ²/ₓ₃
y = 1 - ²/ₓ₃
y - 1 = ⁻²/ₓ₃
x - 1 = -2/y³
y³(x - 1) = -2
y³ = ⁻²/ₓ₋₁
y = ∛⁻²/ₓ₋₁
y = -∛(2x² - 4x + 2)/x - 1
f⁻¹(x) = -∛(2x² - 4x + 2)/x - 1

4 0

3 years ago

I will give you brainliest, if you provide an accurate explanation.

Molodets [167]

Answer:

sqrt(i) =0.707106781 + 0.707106781 i

Most of the numbers we know, and work with, are Real Numbers. The Real Number System includes counting numbers, fractions, terminating decimals, positive numbers, negative numbers, zero, repeating decimals, never ending and non-repeating decimals, numbers that are expressed as radicals, and even pi (π).

The natural numbers are the set of counting numbers

• There are infinitely many numbers in a set of numbers.

• The natural numbers are "closed" under addition and multiplication.

The addition of two natural numbers creates another natural number.

The multiplication of two natural numbers creates another natural number.

closed under addition and multiplication.

BUT ...

The subtraction of two natural numbers does NOT necessarily create another natural number

The division of two natural numbers does NOT necessarily create another natural number

The whole numbers are the set of counting numbers (natural numbers) along with zero

• There are infinitely many numbers in this set of numbers.

• The set of whole numbers is "closed" under addition and multiplication.

Integers:

• The integers are the set of all of the natural numbers,

plus their additive inverses and zero

• The integers are "closed" under addition, multiplication and subtraction,

but NOT under division

Rational Numbers:

• The rational numbers are the set of numbers which can be expressed as a ratio

(a fraction) between two integers.

• Integers are rational numbers since 5 can be written as the fraction 5/1.

• Decimals which terminate are rational numbers.

• Decimals which have a repeating pattern are rational numbers. 1/3 = 0.3333333...

• The rational numbers are "closed" under addition, subtraction, and multiplication. Under division, we run into the problem of division by 0, which makes the statement that "the rationals are closed under division" false. Some texts state that "the rationals are closed under division as long as the division is not by zero" which is a true statement.

Irrational Numbers:

The irrational numbers are the set of number which can NOT be written as a ratio (fraction).

• Decimals which never end nor repeat are irrational numbers.

• Irrational numbers are "not closed" under addition, subtraction, multiplication or division.

• Examples of irrational numbers: rad2, π

8 0

3 years ago

Read 2 more answers

I don’t understand this can someone explain this to me?

olga2289 [7]

Answer:

16x 144

Step-by-step explanation:

4x times 4x then 12 times 12 than that is you answer beacons you can ad 12 and 4x because the 12 has no x

7 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Isamasonip invested $2.000 in

alexgriva [62]

Answer:

Compound Interest= $3440

Final Amount= $5440

Step-by-step explanation:

Compound Interest's formula= P(1+r/100)^n

where P is Principal, r is Rate then n is Years

So in this case,,2000(1+.72)^1 = $3440

Hence, Total Amount = Principal+Interest

therefore T.A= $2000+$3440

T.A=$5440

Thanks... Subjected to Review

7 0

3 years ago