<img src="https://tex.z-dn.net/?f=%20-%201%20%2B%20%20-%202" id="TexFormula1" title=" - 1 + - 2" alt=" - 1 +

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Anarel [89]

2 years ago

ddle" class="latex-formula">

Mathematics

1 answer:

MissTica2 years ago

6 0

Answer:

-3

Step-by-step explanation:

Basically it's like if you owe a dollar (-1), and then you owe 2 more dollars (-2), now you owe 3 dollars (-3).

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Which value, when placed in the box, would result in a system of equations with infinitly many solutions? Y=2x - 5 2y-4x=

Nezavi [6.7K]

Given:

The system of equations is

$y=2x-5$

$2y-4x=?$

To find:

The missing value for which the given system of equations have infinitely many solutions.

Solution:

Let the missing value be k.

We have,

$y=2x-5$

$2y-4x=k$

Taking all the terms on the left side, the given equations can be rewritten as

$-2x+y+5=0$

$-4x+2y-k=0$

The system of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinitely many solutions if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

We have,

$a_1=-2,b_1=1,c_1=5$

$a_2=-4,b_2=2,c_2=-k$

Now,

$\dfrac{-2}{-4}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{5}{-k}$

On cross multiplication, we get

$-k=10$

$k=-10$

Therefore, the missing value is -10.

6 0

2 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

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Step-by-step explanation:

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