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RideAnS [48]
3 years ago
8

Use the unit circle to find the inverse function value in degrees . sin

Mathematics
1 answer:
Deffense [45]3 years ago
5 0
You didn’t provide an image
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Y = -3 x + 9 <br><br> Y= x -9<br><br> Y = 1/3x + 3 <br><br> Y = -3x -9
sashaice [31]

the correct answer is a

8 0
3 years ago
Which could NOT be a view of one face of a triangular prism
allochka39001 [22]

Answer:

D (or the last one) it has multiple faces upon one face do it doesn't count. it also isn't a triangle it's a pentagon.

Step-by-step explanation:

big brain time

7 0
3 years ago
A grain of rice average weight is about 25 milligrams or 0.025 grams. Which of the following shows the weight in grams in scient
zavuch27 [327]

Answer:

2.5x10⁻2

Step-by-step explanation:

4 0
3 years ago
Write the equation of a line that is perpendicular to y=-0.3x+6 and that passes through the point (3,-8)
MariettaO [177]
Hey there!

First, let's look at what perpendicular means. Imagine a cross, where there's all 90 degree angles. That's exactly what we're talking about when we say perpendicular. The given equation is in slope-intercept form, where we have:

y = mx + b
where m is the slope and b is the y-intercept.

When we're writing an equation with a perpendicular slope, we use the negative reciprocal of the given slope. Thus, we can make 0.3 1/3, and take the reciprocal to make 3, and make it negative 3 as it's the negative reciprocal. Now, we know we have a line with the slope of -3 and goes through (-3, 8). We can use the x and y values in this set of points, along with the slope, to create an equation to solve for b. That gives us:

8 = -3(-3) + b
8 = -9 + b
17 = b

Now, since we have slope and y-intercept, we can write our equation as:

y = -3x + 17

Hope this helps!
5 0
3 years ago
Read 2 more answers
Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hi
natta225 [31]
Parameterize the part of the paraboloid within the cylinder - I'll call it S - by

\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle

with 0\le u\le2 and 0\le v\le2\pi. The region's area is given by the surface integral

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv
=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du

Take w=1+36u^2 so that \mathrm dw=72u\,\mathrm du, and the integral becomes

=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw
=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}
=\dfrac\pi{54}(145^{3/2}-1)\approx101.522
7 0
3 years ago
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