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3 years ago

What degrees is Angle c

Mathematics

1 answer:

galben [10]3 years ago

5 0

Answer:

use the protector use it you will get the answer

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8 ^ x * 16 ^ x - 1 equals 1

Lana71 [14]

Answer:

${8}^{x} \times {16}^{x } - 1 = 1 \\ {2}^{3x} \times {2}^{4x} = 1 + 1 \\ {2}^{7x} = {2}^{1} \\ 7x = 1 \\ x = \frac{1}{7}$

7 0

2 years ago

Simplify ( √ 5 ) ( 3 √ 5 )

Amiraneli [1.4K]

Answer:

$5^\frac{5}{6}$

Step-by-step explanation:

$\sqrt{5}$ can be represented as $5^{\frac{1}{2}}$ and $\sqrt[3]5}$ can be represented as $5^{\frac{1}{3}}$ . Therefore, the expression can be rewritten as:

$5^{\frac{1}{2}}*5^{\frac{1}{3}}$

The rule for multiplying two exponents with the same base is you add the exponents. For example: $x^m*x^k=x^{m+k}$

We can use the same property to get:

$5^{\frac{1}{2} +\frac{1}{3}}$

which is just $5^\frac{5}{6}$ after you add the fractions

8 0

2 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$