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Hatshy [7]
4 years ago
13

Solve for x I can’t seem to get it I feel so stupid I’m sorry but please someone help me

Mathematics
2 answers:
photoshop1234 [79]4 years ago
7 0

Use the Side Splitter Theorem (also known as the Basic Proportionality Theorem)

5/3 = 6/x

5x = 3 * 6

5x = 18

x = 3.6

Answer: B. 3.6

kow [346]4 years ago
6 0

In the given triangle the lines are parallel.We can apply Triangle proportionality theorem.

The theorem states: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

We can form a proportion using this theorem :

\frac{5}{3} =\frac{6}{x}

To solve x we cross multiply:

5x=18

Dividing both sides by 5 we get

x=3.6

Second option.

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Answer please!! :) ​
Ivenika [448]
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7 0
3 years ago
A fluid has density 810 kg/m3 and flows with velocity v = z i + y2 j + x2 k, where x, y, and z are measured in meters and the co
mixas84 [53]

Denote the cylindrical surface by S, and its interior by R. By the divergence theorem, the integral of \vec v across S (the outward flow of the fluid) is equal to the integral of the divergence of \vec v over the space it contains, R:

\displaystyle\iint_S\vec v\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec v)\,\mathrm dV

The given velocity vector has divergence

\vec v=z\,\vec\imath+y^2\,\vec\jmath+x^2\,\vec k\implies\nabla\cdot\vec v=\dfrac{\partial(z)}{\partial x}+\dfrac{\partial(y^2)}{\partial y}+\dfrac{\partial(x^2)}{\partial z}=2y

Then the total outward flow is

\displaystyle\iiint_R2y\,\mathrm dV

Converting to cylindrical coordinates gives the integral

\displaystyle2\int_0^{2\pi}\int_0^3\int_0^4r^2\sin\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{0}

6 0
3 years ago
A bar that turns on a fixed point is called a
Sladkaya [172]
I think it's called the fulcrum. <span>The fulcrum is a pivot </span>point<span>.

</span>
8 0
4 years ago
Read 2 more answers
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
What is mean BIJUS ?​
sergeinik [125]

Answer:

abbreviation, shorthand or slang term: BIJS.

( I hope this was helpful) >;D

8 0
3 years ago
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