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3 years ago

0.05 is 1/10 of what

Mathematics

1 answer:

Gennadij [26K]3 years ago

4 0

Answer:

Step-by-step explanation:

.5 times 1/10 is .05. move the decimal over!

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Consider this system of equations. x + 2y = 5, –x + 3y = 6 The x- and y-intercepts of the first equation are found below. x-in

vichka [17]

Answer:

Part 1) The x-intercept is the point (-6,0)

Part 2) The y-intercept is the point (0,2)

Step-by-step explanation:

we know that

The x-intercept is the value of x when the value of y is equal to zero

The y-intercept is the value of y when the value of x is equal to zero

we have

$-x+3y=6$

Part 1) Find the x-intercept

For y=0

substitute the value of y in the linear equation and solve for x

$-x+3(0)=6$

$-x=6$

$x=-6$

The x-intercept is the point (-6,0)

Part 2) Find the y-intercept

For x=0

substitute the value of x in the linear equation and solve for y

$-(0)+3y=6$

$3y=6$

$y=2$

The y-intercept is the point (0,2)

7 0

3 years ago

Read 2 more answers

Choose the best estimate for the mass of a raisin.

sammy [17]

Answer:

A) 0.2g

Step-by-step explanation:

${.}$

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2 years ago

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Solve: 3/K+3 = 8/k-2

fomenos

Answer:

k = 1

Step-by-step explanation:

I will upload a picture with all of my work.

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

An angle's complement is 35°. Find the measure of its supplement. A) 55° B) 65° C) 125° D) 155° E) 180°

kenny6666 [7]

Answer:

C) 125°

Step-by-step explanation:

Supplementary angles add to 180 degrees. Complementary angles add to 90

An angle's complement is 35°.

35+x =90

Subtract 35 from each side

35 -35+x = 90-35

x = 55

We want to find the supplement of this angle

55+y = 180

Subtract 55 from each side

55-55+y = 180-55

y= 125

The angle is 55 so its supplement is 125

7 0

3 years ago