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1 year ago

Use the formula R=3V+2T to find V when R=40 and T=5

Mathematics

2 answers:

steposvetlana [31]1 year ago

8 0

Answer:

V = 10

Step-by-step explanation:

to find V, you must make the equation V = ......

original equation is

R = 3V + 2T

first subtract 2T from both sides

R - 2T = 3V

now divide by 3

V = $\frac{R - 2T}{3}$

now plug in the numbers for R= 40 and T= 5

V = (40 - 2(5)) / 3

V = 30 / 3 = 10

serg [7]1 year ago

3 0

Answer:

V = 10

Step-by-step explanation:

Plug in give information

40 = 3V + 2×5

40 = 3V + 10 we want to get V by itself so subtract 10 from both sides

30 = 3V Divide both sides by 3

10 = V

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2 years ago

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2 years ago

A rectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the s

andre [41]

Answer:

$x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)$

where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation:

Let

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio,

Length of the pool (with the patio)
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio)
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
Area of the pool (w/o the patio) = lw

Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
= (l + 2x)(w + 2x)
= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0 (1)

Let

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

Then using quadratic formula, the roots of the equation 4x² + 2x(l + w) - lw = 0 is given by

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\ = \frac{-2(l + w) \pm \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l + w)^2) + 16lw}}{8} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 6lw + w^2)}}{8}$
$= \frac{-2(l + w) \pm 2\sqrt{l^2 + 6lw + w^2}}{8} \\= \frac{2}{8}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\x = \frac{1}{4}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right) \text{ or }}
\\\boxed{x = -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

Since $(l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \ 0$ , $-\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right)$ is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by

$\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

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2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E3_t%7B%7Cx%2B2%7C%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limi

photoshop1234 [79]

$\displaystyle\int_{-3}^3|x+2|\,\mathrm dx$

Recall the definition of absolute value:

$|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x$

So we can split up the integration interval at $x=-2$ and apply this definition to rewrite the integral as

$\displaystyle\int_{-3}^{-2}(-(x+2))\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx$
$=\displaystyle-\int_{-3}^{-2}(x+2)\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx$
$=-\left(\dfrac12x^2+2x\right)\bigg|_{x=-3}^{x=-2}+\left(\dfrac12x^2+2x\right)\bigg|_{x=-2}^{x=3}$
$=\dfrac12+\dfrac{25}2=13$

7 0

2 years ago