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3 years ago

Divide.write the remainder as a fraction. 125/8

Mathematics

1 answer:

Natali [406]3 years ago

5 0

15 $\frac{5}{8}$ . 125 divided by 8 is 15.625,0,625 times 8 is 5, add 15 and 5/8 and get 15 $\frac{5}{8}$ .

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Can anyone help me asap plzz.

alexandr1967 [171]

the answer is c.

you have to multiply by -3 on both sides to get c on its own.

therefore, c is greater than or equal to -9.

in a, c is less than or equal to 9 so thats not the answer.

In b, c is greater than or equal to 9 but it has to be -9.

in d, c is less than or equal to -9.

the answer is c

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3 years ago

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A tutor works with a group of students. The tutor charges $40 plus $30 for each student in the group. Today the tutor has s stud

IgorLugansk [536]

40 + 30s = 220

30 for each student (s) and the one charge of 40 would get the tutor 220 dollars.

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4 years ago

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If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each o

Ludmilka [50]

Answer:

16/33

Step-by-step explanation:

12C4(total possible outcomes)=495

Since there are 6 married couples there must be 6*[2C2×10C2]=270

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3 years ago

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!

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4 years ago

What is the solution to this equation -4x + 2=26

wolverine [178]

$-4x+2=26$ <em>subtract 2 from both sides</em>

$-4x=24$ <em>divide both sides by (-4)</em>

$\boxed{x=-6}$

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3 years ago

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