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3 years ago

List all the factors of 90 from least to greatest

Mathematics

2 answers:

yanalaym [24]3 years ago

7 0

The factors of 90 include: 1,2,3,5,6,9,10,15,18,30,45,90.
Hope this helps! :)

yawa3891 [41]3 years ago

3 0

1*90
2*45
3*30
4*22.5
5*18
6*15
7*12.8571428571
8*11.25
9*10

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On a coordinate plane, 2 lines are shown. Line P Q has points (negative 8, 2) and (4, 2). Line M N has points (8, 6) and (8, neg

DaniilM [7]

Answer:

Slope of PQ = 0

Slope of MN = infinity

PQ and MN are perpendicular to each other

Step-by-step explanation:

for any two points (x1, y1), (x2, y2)given in coordinate plane slope is given by

$y1 - y2/x1-x2\\\\For \ line \ PQ\\slope = 2 - 2/-8-4 = 0\\\\For \ line \ MN \\slope = 6 - (-8)/8-8 = 1/0 = infinity\\\\$

For any line if slope is zero it is parallel to X axis and perpendicular to Y axis

For any line if slope is infinity it is parallel to Y axis and perpendicular to X axis

Also we know X and Y are perpendicular to each other.

Since slope of PQ is zero it is parallel to X axis and perpendicular to Y axis

Since slope of MN is infinity it is parallel to Y axis and perpendicular to X axis.

Thus two lines PQ and MN are perpendicular to each other.

4 0

3 years ago

Read 2 more answers

Vertical angles are opposite angles with the same vertex <br> true or false

Sati [7]

True
<span>Vertical angles are opposite angles with the same vertex </span>

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3 years ago

Mia has 12 boxes of cookies. Each box contains 36 cookies. she has another 258 cookies in the pantry. How many cookies does she

lana66690 [7]

690

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8 0

2 years ago

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

Sam invests $6000 in two different accounts. The first account paid 12 %, the second account paid 7 % in interest. At the end of

Lunna [17]

Answer:

The amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600

Step-by-step explanation:

Let x be the amount in the account at 12% interest

So, 6000-x is the amount in the second account at 7% interest

$SI = \frac{P \times T \times R}{100}$

First account: $SI=\frac{x \times 1 \times 12}{100}$

Second account : $SI =\frac{(6000-x) \times 1 \times 7}{100}$

We are given that At the end of the first year he had earned $590 in interest.

So, $\frac{x \times 1 \times 12}{100}+\frac{(6000-x) \times 1 \times 7}{100}=590\\x=3400$

So,the amount in the account at 12% interest is $3400

The amount in the second account at 7% interest =6000-x=6000-3400=2600

Hence the amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600

8 0

3 years ago