All categories

3 years ago

Right angle triangle problem solve for x

Mathematics

1 answer:

vladimir2022 [97]3 years ago

5 0

Since the side labeled 3 is 1/3 of 9 it's safe to assume the angle opposite is 1/3 of 90 degrees, making it 30 degrees. Here we are gonna use sin. The formula is gonna be sin(30)=3/x which can be turned into x(sin(30))=3 then divide both sides by sin(30) getting x=6. For Y, you're going to do Pythagorean Theorem. First, take 6^2 and subtract 3^2 which is 25 then square root that to get 5 which is the height of the smaller triangle. Then Pythagorean theorem again, 5^2 plus 9^2 equals 106 square root that to get roughly 10.3. Getting you the answeres of X=6 and Y=10.3

You might be interested in

6-3/8 I need help please

Daniel [21]

Answer:

5.625

Explanation:

3 0

3 years ago

Read 2 more answers

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

What is the area of triangle ABC?

Ann [662]

Answer:

A = 3 units^2

Step-by-step explanation:

The area of a triangle is

A = 1/2 bh where b is the length of the base and h is the height

b = the distance from B to C = 4 units = 2 ( watch the scale)

h = the distance from A to C along the x axis =6 units = 3 (watch the scale)

A = 1/2 (2)(3)

A = 3 units^2

7 0

3 years ago

Read 2 more answers

Simplify.Your answer should contain only positive exponents.(x^3y^-2)-4

Galina-37 [17]

X^3(1/y^2)-4 is (x^3y^-2)-4 simplified with only positive exponents

8 0

3 years ago

PRECALC HELP<br> please help with the image above (translation up and down)

garik1379 [7]

Answer:

fourth option

Step-by-step explanation:

Given f(x) then f(x + a) represents a horizontal translation of f(x)

• If a > 0 then a shift left of a units

• If a < 0 then a shift right of a units

Thus

f(x) = (x - 11)³ ← has been translated right by 11 units

Given f(x) then f(x) + c represents a vertical translation of f(x)

• If c > 0 then a shift up of c units

• If c < 0 then a shift down of c units

Thus

f(x) = (x - 11)³ + 4

represents a translation 11 units right and 4 units up

7 0

3 years ago