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2 years ago

14. Which makes the number sentence true? 12 - 3= 7+_

Mathematics

1 answer:

stepan [7]2 years ago

8 0

The answer is 2 in order to make the sentence true

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What is the solution to the equation |x − 4| = 17?

andrey2020 [161]

$~~~~~~|x-4| = 17\\\\\implies x -4 = 17~~~~ \text{or}~~~~~ x -4 = -17\\\\\implies x = 17+4~~~~\text{or}~~~~~ x = -17+4\\\\\implies x = 21~~~~~~~~~\text{or}~~~~~x = -13$

8 0

1 year ago

Consider m = y2 - y1/ x2 - x1 . Which x1 and x2-values would determine that the line is vertical? Justify your answer

Y_Kistochka [10]

Answer:

$x_2=x_1$

Step-by-step explanation:

We were given the slope formula;

$m=\frac{y_2-y_1}{x_2-x_1}$

This line is vertical if the denominator is zero.

That is when $x_2-x-1=0$

This implies that;

$x_2=x_1$

Justification;

When $x_2=x_1$ , then, the line passes through;

$(x_1,y_1)$ and $(x_1,y_2)$

The slope now become

$m=\frac{y_2-y_1}{x_1-x_1}=\frac{y_2-y_1}{0}$

The equation of the line is

$y-y_1=\frac{y_2-y_1}{0}(x-x_1)$

This implies that;

$0(y-y_1)=(y_2-y_1)(x-x_1)$

$0=(y_2-y_1)(x-x_1)$

$\frac{0}{y_2-y_1}=(x-x_1)$

$0=(x-x_1)$

$x=x_1$ ... This is the equation of a vertical line.

3 0

2 years ago

Find an equation of a line that comes close to the points listed in the table.

sergey [27]

$y=x+2\\x=6\to y=6+2=8\\x=10\to y=10+2=12\\x=12\to y=12+2=14\\x=16\to y=16+2=18\\x=18\to y=18+2=20\\-----------------------\\y=x+3\\x=6\to y=6+3=9\\x=10\to y=10+3=13\\x=12\to y=12+3=15\\x=16\to y=16+3=19\\x=18\to y=18+3=21\\-----------------------\\Answer:y=x+1.5$

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2 years ago

Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

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3 years ago

A linear function is graphed below.

Evgen [1.6K]

Answer:

$-1\frac{1}{2}$