All categories

2 years ago

Help help help math math

Mathematics

2 answers:

VashaNatasha [74]2 years ago

8 0

2|x-3|=30
|x-3|=15
x-3=-15
x-3=15
x=-12
x=18

Contact [7]2 years ago

4 0

$\frac{2 |x - 3| }{5} = 6 \\$

$5 \times \frac{2 |x - 3| }{5} = 5 \times 6 \\$

$2 |x - 3| = 30$

$\frac{2 |x - 3| }{2} = \frac{30}{2} \\$

$|x - 3| = 15$

$x - 3 = 15$

$x - 3 + 3 = 15 + 3$

$x = 18$

$x - 3 = - 15$

$x - 3 + 3 = - 15 + 3$

$x = - 12$

You might be interested in

Waht is 2t2 math fir math clas

Lelechka [254]

2+2 is 4 LOL you just gotta add 1+1+1+1

4 0

3 years ago

Read 2 more answers

Find the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.

riadik2000 [5.3K]

<span>First of all to calculate the distance between two points we can use distance formula

d=Square Root [(x2-x1)^2 + (y2-y1)^2]

Now substitute the given points p(x1,y1) and q(x2,y2)in above distance formula

The values are X2=3, X1=8and Y2=8and Y1=2.

After Substituting the values

d=Square Root[(-5)^2+(6)^2]

d=Square Root(25+36]

d=Square Root[61]

d=7.8

7.8 is the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.</span>

7 0

3 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

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

8 0

3 years ago

PLEASE HELP ME WITH THIS THANKS

Lady bird [3.3K]

What is M when R = 200? *

Step-by-step explanation:

6 0

3 years ago

The graph of the function f(x)=−4(x−3)2+9 will be a parabola opening in which direction?

sesenic [268]

Answer:

downwards

Step-by-step explanation:

the equation of a parabola in vertex form is

f(x) = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

• if a > 0 then parabola opens upwards

• if a < 0 then parabola opens downwards

for

f(x) = - 4(x - 3)² + 9 with a = - 4 < 0 , then

the parabola opens downwards

8 0

2 years ago