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3 years ago

The sum of two numbers is 25. Their difference is 9. What are the two numbers?

2 answers:

5 0

Using x and y as the numbers

x+y=25 -------#1
x-y=9 -------#2

Using addition of #1 and #2
x+y+x-y= 25+9
2x+0y=34
2x=34
x=17

Using the value of x and plugging said value into #1 or #2, we can find y
x=17

x+y=25
17+y=25
y=8

OR

x-y=9
17-y=9
-y=-8
y=8

The numbers are 8 and 17

Julli [10]3 years ago

3 0

I'm Just guessing 16 and 9.

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Fill in the blanks below in order to justify whether or not the mapping shown

MrRa [10]

Answer:

wheres the picture?

Step-by-step explanation:

6 0

2 years ago

Only answer if you actually want to help me please

mamaluj [8]

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5 0

3 years ago

Find the length of UC on the attached diagram. Thank you!

ValentinkaMS [17]

<h2>Answer </h2>

The length of UC is 18

<h2>Explanation </h2>

First we are going to find the length of JN; then we are subtracting from it the length of JU plus the length of CN.

We can infer from our picture that JN is 82 + 105, so JN = 187

We can also infer that JU = JH + HU

JU = 22 + 96

JU = 118

We can also infer that CN = 51

Now we can fin the length of UC:

$UC=JN-(JU+CN)$

$UC=187-(118+51)$

$UC=18$

We can conclude that the length of UC is 18.

3 0

3 years ago

Read 2 more answers

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

2 years ago

What expression is equivalent to 1/36

IrinaVladis [17]

The expression equivalent to 1/36 are :-

$\frac{1}{36} \: = \: \frac{2}{72} ; \frac{3}{108} ; \frac{4}{144} ; \frac{5}{180}$

<h2 />

3 0

2 years ago