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3 years ago

The number 132 can be written in the form

1 answer:

3 0

$132|2 \\ 66 \ | 2 \\ 33 \ |3 \\ 11 \ |11 \\ 1 \\\\ 132=2*2*3*11 \\\\ 132=2p*q*r \\\\ 2p=2*2 \to\boxed{p=2} \\\\ \boxed{q=3} \ \ \ \ \ \ \ \ \ \ \ \ \boxed{r=11}$

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Grade 8 math PLEASE HELP

jonny [76]

Answer:

Step-by-step explanation:

If each input value leads to only one output value, classify the relationship as a function.

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3 years ago

Slope = 4, passing through (-6, 8)

Mademuasel [1]

The point slope form is y - 8 = 4(x + 6) and the slope intercept form would be y = 4x + 32.

In order to find this, we'll start with the base form of point-slope form.

y - y1 = m(x - x1) ----> Plug in the numbers

y - 8 = 4(x - -6) ----> Simplify

y - 8 = 4(x + 6)

Now to find slope intercept form, solve for y.

y - 8 = 4(x + 6) ----> distribute the 4

y - 8 = 4x + 24 -----> add 8 to both sides

y = 4x + 32

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3 years ago

A city known for its temperature extremes started the day at -5 degrees Fahrenheit. The temperature increased by 78 degrees Fahr

charle [14.2K]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Starting temperature = - 5°F

Temperature at midday =increase in temperature by 78°F = (starting temperature + 78°F)

Temperature at nightfall = decrease in temperature by 32°F = ( temperature at midday - 32°F)

Temperature at midday:

(-5°F + 78°F) = 73°F

Temperature at nightfall:

(Temperature at midday - 32°F)

(73°F - 32°F) = 41°F

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3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

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3 years ago

Not sure how to solve the problem in the pic :)

yulyashka [42]

Answer:

Step-by-step explanation:

do the division first then subtract 11

x/3 - 11

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3 years ago