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3 years ago

15

I am a 2-digit number. I am a am an even number. The sum of all my digits together is 3. The order of my digits goes smallest to

greatest. (Left to right)

2 answers:

julia-pushkina [17]3 years ago

8 0

Answer:

12

Step-by-step explanation:

12 = 2 digits

1 + 2 = 3

1 < 2

pav-90 [236]3 years ago

4 0

Answer:

12

Step-by-step explanation:

...................

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Obtén lo que se te pide, dadas las dos matrices siguientes.

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I think its b because it makes sense

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4 years ago

5th grade math. correct answer will be marked brainliest.

azamat

Answer:

3 1/3 is your answer.

Step-by-step explanation:

Rewriting equation:

4 + 5/6 - 1 - 1/2

4 - 1 = 3

5/6 - 1/2 = ?

Find the LCD of 5/6 and 1/2 and rewrite to solve.

5/6 - 3/6 = 2/6

2/6 = 1/3

3 + 1/3 = 3 1/3

Whole number: 3

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3 years ago

Read 2 more answers

A beam of light from a monochromatic laser shines into a piece of glass. The glass has thickness L and index of refraction n=1.5

boyakko [2]

Additional information to complete the question:

How long does it take for a short pulse of light to travel from one end of the glass to the other?

Express your answer in terms of some or all of the variables f and L. Use the numeric value given for n in the introduction.

T = ___________ s

Answer:

$T = \frac{15}{f}$

Step-by-step explanation:

Given:

Thickness og glass = L

Index of refraction n=1.5

Frequency = f

$Wavelength = \frac{L}{10}$

λ(air) $= \frac{L}{10}$

λ(glass) = λ(air) / n

= $\frac{\frac{L}{10}}{1.5}$

= $\frac{L}{10} * \frac{1}{1.5}$

= $\frac{L}{15}$

V(glass) = fλ(glass)

$= f * \frac{L}{15}$

$T = \frac{L}{V_{glass}} = \frac{15}{f}$

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3 years ago

What is the inverse of the function f(x) = x +3?

harkovskaia [24]

Answer:

x=y+3

Step-by-step explanation:

you just swap variables.

4 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