All categories

1 year ago

What is log Subscript 15 Baseline 2 cubed rewritten using the power property?

Mathematics

2 answers:

makkiz [27]1 year ago

7 0

Answer:

The Answer is D on Edge :)

Step-by-step explanation:

Bas_tet [7]1 year ago

5 0

Answer:

D) 3Log15^2

Step-by-step explanation:

is this right?

You might be interested in

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

2 years ago

Draw 2 squares. The first one should have an area of 12 square units.The second one should have an area that is DOUBLE that of t

statuscvo [17]

Step-by-step explanation:

Given the information:

1st square: 12 square units
2nd square: DOUBLE that of the first square = 2*12 = 24 square units

From that, we can determine the length of the side in each square:

1/ $x^{2} = 12$

<=> x = 2 $\sqrt{3}$

2/ $a^{2} = 24$

<=> a = 2 $\sqrt{6}$

Please have a look at the attached photo.

Hope it will find you well.

6 0

2 years ago

Find the rate of change of the function h(x) = 2 x on the interval 2 ≤ x ≤ 4

dangina [55]

For a linear function, the instantaneous rate of change is everywhere equal to the slope. Thus the rate of change of the function h(x)=2x on the interval 2≤x≤4
The rate of change of the function given will equal to its slope, thus;
slope,m=(y-1-y)/(x_1-x)
=(2*4-2*2)/(4-2)
=(8-4)/2
=4/2
=2

the answer is 2

7 0

2 years ago

What is the domain of the function y= square root x - 5 - 1

IrinaK [193]

Answer:

$x\geq 5$

Step-by-step explanation:

The function that we have to study in this problem is

$y=\sqrt{x-5}-1$

The domain of a function is defined as the set of all the possible values of x that the function can take.

For a square-root function, there are some limitations to the possible value of the argument in the root.

In particular, the argument of a square root must be equal or greater than zero, because the square root of a negative number is not defined.

Therefore, in this case, we have to set the following condition for the domain:

$x-5\geq 0$

And by solving, we get

$x\geq 5$

which means that the domain of this function is all real numbers equal or greater than 5.

5 0

2 years ago

Calculate the flux of the vector field F⃗ (x,y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 wit

ikadub [295]

The square (call it $S$ ) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is $\vec n=\vec\imath-\vec k$ , which is enough information to figure out the equation of the plane containing $S$ :

$(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x$

We can parameterize this surface by

$\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+x\,\vec k$

for $0\le x\le\frac3{\sqrt2}$ and $0\le y\le3$ . Then the flux of $\vec F$ , assumed to be

$\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k$ ,

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}$

3 0

2 years ago