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3 years ago

What is the length of this ruler to the correct degree of precision? 16 mm 1.61 cm 1.61 mm 1.610 cm

Mathematics

2 answers:

ivolga24 [154]3 years ago

8 0

Answer:

If you will look at the red line that is spread on the ruler,if measured with divider , it will lie between 1.6 cm to 1.7 cm.

1 cm =10 mm

Coming to the Options

---The length of the Red Ruler can be 16 mm.

---1.61 cm and 1.610 cm are same length.

---1.61 mm is rejected.

But from this kind of ruler you can't say with exact degree of Precision that length will be either 1.61 cm or 1.610 cm.Because it can't measure 1.61 cm.The scale can measure lengths upto a single decimal place that is upto tenths.

So, the length of this ruler to the correct degree of precision will be = 16 mm =1.6 cm

Option A

zavuch27 [327]3 years ago

5 0

Answer:

16 mm

Step-by-step explanation:

Knowing that the longest lines with the numbers marked on them is the tens, I realized there was only one line the red line crossed. The other 6 was from one little mm and the med-long line.

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Through: (2, 2), parallel to y = x +4

dexar [7]

Answer:

y = x

Step-by-step explanation:

We can use slope-point form.

(y-y1) = m(x-x1)

m is a slope. If the line is parallel than slopes are the same.

y = x + 4, for this line slope = 1, so for line we have to find slope is also 1.

(y-y1) = m(x-x1)

m=1, x1 = 2, y1 = 2

(y - 2) = 1(x-2)

y - 2 = x - 2

y = x

8 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

How many minutes are in m hours?

notka56 [123]

Answer:

the 2nd one (60m minutes)

Step-by-step explanation:

60 minutes in an hour, so if m is the number of hours you would times that by the number of minutes in an hour which is sixty. 60m is 60 times m

8 0

3 years ago

Chicos estoy en un aprieto ayudarme pls Sabiendo que una falda es 5 € más cara que una blusa y que si compro 6 faldas y 4 blusas

poizon [28]

Answer:

El precio de la falda es 50 €.

Step-by-step explanation:

Lo que sabemos:

$F = 5 + B$ (1)

En donde F es por falda y B por blusa.

Si compras 6 faldas y 4 blusas y pagas 480 €:

$6F + 4B = 480$ (2)

Entonces, al introducir la ecuación (1) en (2) podemos hallar el precio de cada falda.

$6F + 4B = 480$

$6(5 + B) + 4B = 480$

$10B = 480 - 30$

$B = \frac{450}{10} = 45$

Finalmente, el precio de la falda se puede encontar con la ecuación (1):

$F = 5 + 45 = 50$

Por lo tanto, el precio de la falda es 50 €.

Espero que te sea de utilidad!

7 0

2 years ago

There are 100 students at a college who have taken a course in calculus, 110 who have taken a course in discrete mathematics, an

vesna_86 [32]

Answer:

110 students

Step-by-step explanation:

The number of students who have only taken calculus is given by the number of students who have taken calculus minus the students who have taken both classes:

$C =100 -50\\C= 50\ students$

The number of students who have only taken discrete mathematics given by the number of students who have taken discrete mathematics minus the students who have taken both classes:

$D =110 -50\\C= 60\ students$

The number of students that have taken a course in either calculus or discrete mathematics is:

$E = C+D = 50+60\\E= 110\ students$

7 0

3 years ago