Help me! Mathematics, quick, I'm on Time Limit!<br><br> Thank you all very much!

Let Angle w be an acute angle. Use a calculator to approximate the me

serious [3.7K]

Answer:

Step-by-step explanation:

Given the trigonometry function as shown sin W = 0.79, to get W, the following steps must be followed.

Step 1: Take the arcsin of both sides of the equation

$sin^{-1}(sinW) = sin^{-1}0.79\\W = sin^{-1}0.79$

Step 2: Use a calculator to get the resulting value of W

$W = 49.46^{0}$

$W = 49.5^{0}$ (to nearest tenth of a degree)

This gives the acute angle needed.

Note that acute angle are angles that are less than 90°

8 0

3 years ago

A $400 suit is marked down to $260. Find the percent of markdown.

notka56 [123]

The answer is B. -$140

4 0

2 years ago

Read 2 more answers

Points and Brainliest Giveaway!<br><br> What is the simplified form?<br> (2x^2 + 3x) + (4x^2 -7)

nika2105 [10]

Answer:

Explanation:

We can factor the numerator and denominator as;

(

x

−

2

)

(

x

−

1

)

2

x

(

x

−

1

)

We can now cancel common term in the numerator and denominator:

(

x

−

2

)

(

x

−

1

)

2

x

(

x

−

1

)

⇒

x

−

2

x

However, we cannot divide by

0

so we must exclude:

2

x

=

0

⇒

x

=

0

and

x

−

1

=

0

⇒

x

1

x

2

−

3

x

+

2

x

2

−

2

x

=

x

−

2

x

Where:

x

≠

0

and

x

≠

1

Or

x

2

−

3

x

+

2

x

2

−

2

x

=

x

2

x

−

2

x

=

1

2

−

1

x

Where:

x

≠

0

and

x

≠

1

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

The equation of a line is y=-2x+1. What is the equation of the line that is parallel to the first line and passes through (2,2)?

Murrr4er [49]

Answer:

The equation in slope-intercept form is y=-2x+6.

The equation in standard form is 2x+y=6.

The equation in point-slope form is y-2=-2(x-2).

Step-by-step explanation:

The slope-intercept form of a line is y=mx+b where m is the slope and b is the y-intercept.

Parallel lines will have the same slope and different y-intercept.

Anyways the slope of y=-2x+1 is -2.

So the equation of our line we are looking for is -2.

So we know our equation is in the form y=-2x+b.

We must inf b using y=-2x+b with (x,y)=(2,2).

y=-2x +b with (x,y)=(2,2)

2=-2(2)+b

2=-4+b

Add 4 on both sides:

2+4=b

Simplify:

6=b

The equation is y=-2x+6.

Now it didn't say what form it wanted.

There are some forms I can give you like standard and point-slope form.

There is also general form but it is not too much different from standard form.

Standard form is ax+by=c where a,b, and c are integers if possible.

Point-slope form is y-y1=m(x-x1) where (x1,y1) is a point on the line and m is the slope.

So let's go for standard form (ax+by=c) first:

y=-2x+6

add 2x on both sides:

2x+y=6

This is standard form because it is in the form

ax+by=c.

Ok we know point (2,2) is on our line and we also know we have slope,m, is -2.

Point-slope form is

y-y1=m(x-x1)

y-2=-2(x-2)

8 0

3 years ago

Help me! Mathematics, quick, I'm on Time Limit! Thank you all very much!