#11 Please answer the question

Solve for x. Round to the nearest tenth of a degree, if necessary.

ziro4ka [17]

Answer:

x ≈ 28.8°

Step-by-step explanation:

Using the sine ratio in the right triangle

sin x = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{3.9}{8.1}$ , then

x = $sin^{-1}$ ( $\frac{3.9}{8.1}$ ) ≈ 28.8° ( to the nearest tenth )

3 0

3 years ago

What are the zeros of f(x)=x^2+x-20?

zzz [600]

F(x) = x² + x - 20 = x² + 5x - 4x - 20 = x(x + 5) - 4(x + 5) = (x + 5)(x - 4)

f(x) = 0 ⇔ (x + 5)(x - 4) = 0 ⇔ x + 5 = 0 or x - 4 = 0 ⇒ x = -5 or x = 4

Answer: C. x = -5 and x = 4.

3 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

g Which of the following is true about a p-value? Group of answer choices It measures the probability that the null hypothesis i

scZoUnD [109]

Answer:

It measures the probability of observing your test statistic, assuming the null hypothesis is true.

Step-by-step explanation:

The p-value, also known as the probability value <u>measures the probability of observing your test statistic, assuming the null hypothesis is true.</u>

A low p-value means a higher chance of the null hypothesis to be true.

It lies between 0 and 1. A small p-value indicates fewer chances of the null hypothesis to be true.

5 0

3 years ago

0.3r=2.1. whats the solution

vivado [14]

In order to find "r", you need to isolate it. So in order to isolate it, you do the inverse operation. In this case, it would be division because 0.3 is being multiplied by "r". So you divide by 0.3 on both sides to get 7.

3 0

3 years ago