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3 years ago

Can someone please help me?! Marking Brainliest!!!!!

Mathematics

2 answers:

irga5000 [103]3 years ago

3 0

Answer:

$5\sqrt{2}$

Step-by-step explanation:

The formula for the length of a vector/line in your case.

$L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{[4 - (-1)]^2 + [2 -(-3)]^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$

aleksklad [387]3 years ago

3 0

Answer: Mark them BRAINLIEST

Step-by-step explanation:

Because they helped my little sister out and they deserve it! :D

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Joe spent half of his weekly allowance playing mini-golf. To earn money, his parents let him weed the garden for $6. What is his

frosja888 [35]

Answer: Joe's weekly allowance = $12

Step-by-step explanation:

Joe spent half of his weekly allowance playing mini-golf

Let a = his weekly allowance

This means Joe spent a/2 playing mini-golf.

To earn money, his parents let him weed the garden for $6

This means Joe's total money

= a + 6

What is his weekly allowance if he ended with $12?

This means he had $12 left after spending a/2 from ta total of (a +6)

Therefore,

(a +6) -a/2 = 12

Taking LCM of 2

[2(a+6)-a]/2=12

Cross multiplying by 2

2a + 12-a = 24

a+12 =24

a = 24-12 =$12

Joe's weekly allowance = $12

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3 years ago

Use the table to identify the values of p and q that should be used to factor x^2+9x-10 as (x+p)(x+q)

likoan [24]

X^2 + 9x -10 = (x -1)(x +10) = (x +(-1)) (x + 10)
p = -1 and q = 10
answer is A.

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Which is a solution to the equation y=5x-6

Vilka [71]

$y=5x-6 \\ \\ the \ graph\ of \ a \ linear \ equation \ is \ a \ straight \ line$

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3 years ago

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Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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zubka84 [21]

Your answer is A, 72°.

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