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3 years ago

PLEASE PLEASE PLEASE HELP

Mathematics

1 answer:

Eduardwww [97]3 years ago

6 0

first make two triangles using M , so there iwll be 90-45-45, and 30-60-90 then use the side rules of those special triangles

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At higher elevations water boils at lower temperatures. The temperature at which water boils is called the boiling point of wate

Leya [2.2K]

B(x) = 100 - 3.5x
84 = 100 - 3.5x
3.5x = 100 - 84 = 16
x = 16/3.5 = 4.6 kilometers

96 = 100 - 3.5x
3.5x = 100 - 96 = 4
x = 4/3.5 = 1.1 kilometers

Therefore, the boiling point of water is between 84°C and 96°C at elevations of between 4.6 kilometers and 1.1 kilometers.

7 0

3 years ago

Mr green makes some compost he mixes soil, manure and leaf mould in the ratio 3:1:2 mr greenmakes 72 literes of the compost. How

love history [14]

Answer:

$36\ liters\ of\ soil$

Step-by-step explanation:

we know that

The ratio of soil, manure and leaf mould is $3:1:2$

That means

For every 6 liters of the compost, he use 3 liters of soil

so by proportion

Find how many liters of soil does he use for 72 liters of compost

$\frac{3}{6}=\frac{x}{72}\\ \\x=3*72/6\\ \\x=36\ liters\ of\ soil$

8 0

3 years ago

Please explain how to get the answer.

Maru [420]

2r=R

Because 2 of "r" make 1 R

3 0

3 years ago

Read 2 more answers

I needddd heelpppp please

Nostrana [21]

Okay I am not 100% sure because that is an extremely confusing question. But here is what I got.

(A) 2c² = 2bc + ac/2 Given

(B) 4c² = 2bc + ac Multiplication and Distribution

(C) 4c = 4b + a Division and Distribution

(D) 4c - a + 4b Subtraction

(E) 4b = 4c - a Symmetric

6 0

4 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

4 years ago