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3 years ago

Find m∠LOM for brainliest

Mathematics

1 answer:

Vlad1618 [11]3 years ago

3 0

Answer:

p = 6

Step-by-step explanation:

4 + 2p = 10( (3/5)p - 2 )

Distribute the 10

4 + 2p = 6p - 20

Subtract 4 from both sides

2p = 6p - 24

Subtract 6p from both sides

-4p = -24

Divide both sides by -4

p = 6

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bonufazy [111]

So
Total cost of the workbooks is d
And the number of workbooks is t
Therefore d = 95
95=19t
t=95/19=5
5 workbooks

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3 years ago

For a LOT OF POINTS answer this question.

klasskru [66]

Answer:

Step 2 has the error

1. The whole number (20) was left out

2. Adding different signs, we subtract and retain the sign from the larger number

Correction:

Step 2: 20 - 17x = 34x + 60

Step 3: -17x -34x = 60 - 20

Step 4: -51x = 40

Step 5: x = 40/-51

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3 years ago

Read 2 more answers

Repeated addition:<br> [ 1,3] + [1,3]+[1,3]= [a,b]<br> A=? B=?

sammy [17]

Answer:

I THINK a=3 b=9

Step-by-step explanation:

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2 years ago

3.5821568896×10−10 What does this equal?

lana [24]

Answer:

25.821568896

Step-by-step explanation:

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2 years ago

Read 2 more answers

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago