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3 years ago

Help ASAP please I need the answers please please

Mathematics

1 answer:

max2010maxim [7]3 years ago

5 0

Answer:

1 is adjacent and linear pair 2 is adjacent and not linear pair 3 is not adjacent and not linear pair

Step-by-step explanation:

4) angle 4and5 are adjacent

5)angle 1and4 are linear pair

6)angle 2and3 are linear pair

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Explain what the graph of the function represents. Be sure to use complete sentences. f(x) = -2/3 + 490.

choli [55]

Answer:

parallel functions.

lines negative.

functions different in that they do not have the same y-intercept or x-intercept.

Step-by-step explanation:

from brainly

itisdougthepugs

1. y = -⅔x + 490

2. I would start with the y-intercept(490) on the graph. Then from that point, I would use rise/run to find the next point: I would go down two points (because the slope is negative), and run three points to the right.

3. f(x) = -⅔ + 490. The graph represents the total profits of last month.

4. View file below.

5. The graphs of these functions are similar in the fact that they are parallel to each other, therefore having the same slope.Both lines are also negative. They are different in that they do not have the same y-intercept or x-intercept.

8 0

3 years ago

For the differential equation 3x^2y''+2xy'+x^2y=0 show that the point x = 0 is a regular singular point (either by using the lim

Svetlanka [38]

Given an ODE of the form

$y''(x)+p(x)y'(x)+q(x)y(x)=f(x)$

a regular singular point $x=c$ is one such that $p(x)$ or $q(x)$ diverge as $x\to c$ , but the limits of $(x-c)p(x)$ and $(x-c)^2q(x)$ as $x\to c$ exist.

We have for $x\neq0$ ,

$3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0$

and as $x\to0$ , we have $x\cdot\dfrac2{3x}\to\dfrac23$ and $x^2\cdot\dfrac13\to0$ , so indeed $x=0$ is a regular singular point.

We then look for a series solution about the regular singular point $x=0$ of the form

$y=\displaystyle\sum_{n\ge0}a_nx^{n+k}$

Substituting into the ODE gives

$\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0$

$\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k$
$\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k$
$\displaystyle+\sum_{n\ge2}a_{n-2}x^{n+k}=0$

From this we find the indicial equation to be

$(3(k-1)+2)ka_0=0\implies k=0,\,k=\dfrac13$

Taking $k=\dfrac13$ , and in the $x^{k+1}$ term above we find $a_1=0$ . So we have

$\begin{cases}a_0=1\\a_1=0\\\\a_n=-\dfrac{a_{n-2}}{n(3n+1)}&\text{for }n\ge2\end{cases}$

Since $a_1=0$ , all coefficients with an odd index will also vanish.

So the first three terms of the series expansion of this solution are

$\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}$

with $a_0=1$ , $a_2=-\dfrac1{14}$ , and $a_4=\dfrac1{728}$ .

6 0

4 years ago

Very urgent!!!

LekaFEV [45]

Answer:

Step-by-step explanation:

Let volume of container = x

1/6x + 5 = 1/2x

x/6 + 5 = x/2

Multiply through by 6

x + 30 = 3x

Subtract 30 from both sides

x + 30 - 30 = 3x - 30

x = 3x - 30

x - 3x = - 30

-2x = - 30

x = 30 /2

x = 15

4 0

3 years ago

A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 11 cm long and 4 cm wide. The di

marin [14]

Figure=1rectangle+2 semicircles
1semicircle=1/2circle
2 (1/2 circles)=1 circle
figure=1 rectangle+1 circle
area of rectangle=legnth times width
area of circle=pir^2

so the width of the rectangle=diameter of the circle
widht=4

diameter=2r
d/2=r
width=d=4
4/2=r=2

area of circle=3.14 times 2^2
are of circle=3.14 times 4
aera of circle=12.56

area of rectangle=11 times 4
area=44

total=add them up
12.56+44=56.56 cm^2

8 0

3 years ago

Guys please I need help please!!!!

nasty-shy [4]

Answer:

Need Help With What Question?

Step-by-step explanation:

What Question you need help with?

3 0

3 years ago