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3 years ago

ANSWER ASAP 10 POINTS

Mathematics

2 answers:

Sav [38]3 years ago

7 0

x + 33° = 180° (It's forming a linear pair that's why the sum is 180°.)

x = 180° - 33°

x = 147°

Marianna [84]3 years ago

5 0

Other person is right! writing here so u can give brainliest:)

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Somebody help me with this math plz!!!

Leya [2.2K]

Answer:

-60 miles per hour

Step-by-step explanation:

3 0

3 years ago

Write an equation for a rational function with the given characteristics.

inessss [21]

The vertical asymptote requires a factor of (x-4) in the denominator.

The double zero requires a factor of (x-1)² in the numerator.

At x=0, these factors will give a value of (-1)²/(-4) = -1/4, so another constant factor of -12 is needed to make the y-intercept be 3.

Your rational function is ...

... y = -12(x -1)²/(x -4)

8 0

3 years ago

Identify the solution set of 3 In 4 = 2 In x. 0 {6} O {-8, 8) O {8}

Aleksandr-060686 [28]

Answer:

x = 8

Step-by-step explanation:

3 ln 4 = 2 ln x ( x > 0 )

ln $4^{3}$ = ln $x^{2}$

x² = 4³

x² = (2³)²

x = ± 2³

$x_{1}$ = 8

$x_{2}$ = - 8 ( strange root, does not satisfy the domain of given function )

x = 8

3 0

3 years ago

hey how do i solve this problem i’m clueless and don’t know how to do it. if you can could you explain it please?

Kamila [148]

(6g^4h^5)^2. We have to square each term individually. Remember, squaring an exponent is just multiplying the exponents together.

6^2 = 36

(g^4)^2 = g^8

(h^5)^2= h^10

The final answer: 36g^8h^10

6 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

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

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago