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3 years ago

In a candy mix there are 4 green bars for every 3 red bars how many red bars are ther if there are 200 green bars

Mathematics

2 answers:

nexus9112 [7]3 years ago

5 0

200 ÷ 4 = 50 (4 bars, 200 total)
50 x 3 = 150 (Above answer times 3 red)
So if there are 200 greens and every 4 green there's 3 red, then there would be
150.

puteri [66]3 years ago

3 0

Set up a proportion.
Solve for x

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2.5 quarters weigh approximately 1 ounce. How many pounds would you expect a

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1.5 pounds

Step-by-step explanation:

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60/40 = 1.5 pounds

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3 years ago

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2 years ago

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Identify the equation of the linear function through the points (- 2,- 7) and (1,2). Then state the rate of change of the

Assoli18 [71]

Answer:

The equation of line is y = 3 x - 1 , i.e option A .

The rate of change = m = 3

Step-by-step explanation:

Given as :

The points are

( $x_1$ , $y_1$ ) = (- 2, -7)

( $x_2$ , $y_2$ ) = (1 , 2)

Let The slope = m

Now, The slope is calculated in points form

So, Slope = $\dfrac{y_2-y_1}{x_2-x_1}$

I.e m = $\dfrac{2-(-7)}{1-(-2)}$

or, m = $\dfrac{9}{3}$

I.e m = 3

So, The slope of points = m = 3

I,e Rate of change = m = 3

Now, the equation of line in slope - point form can be written as

y - $y_1$ = m ( x - $x_1$ )

where m is the slope of line

i.e y - (-7) = ( 3 ) × ( x - (-2) )

or, y + 7 = 3 × (x + 2)

∴ y + 7 = 3 x + 6

Or, y = 3 x + 6 - 7

Or, y = 3 x - 1

So, The equation of line is y = 3 x - 1

Hence, The equation of line is y = 3 x - 1 , i.e option A .

And The rate of change = m = 3 Answer

8 0

3 years ago

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

3 years ago