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3 years ago

Is the answer A or B or C or D??

Mathematics

2 answers:

Mice21 [21]3 years ago

6 0

Answer:

Step-by-step explanation:

(negative numbers) - (positive numbers) = (negative numbers) + -(positive numbers)

Bye!

maxonik [38]3 years ago

3 0

Answer:

Step-by-step explanation:

$-3\frac{1}{8}+(-2\frac{1}{2})=\\ \\=-3\frac{1}{8}-2\frac{1}{2}\\ \\(+)(-)=(-)$

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In triangle VWX, side VW and side XV are congruent and the measure of X is 19 degrees. What is the measure of angle V?

Alchen [17]

Answer:

80.5°

Step-by-step explanation:

Every triangle is measured to be 180°. Subtract X which is 19 from 180

180-19=161

Because V and W are congruent, they are the same length. So you divide 161 by w to get 80.5

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3 years ago

What comes after 7 1/4 on a number line

djverab [1.8K]

If you’re looking for the next whole number, the answer would be 8 if you’re looking for the next number in this phrase the answer would be 7 2/4

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3 years ago

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A research scientist reports that mice will live an average of 40 months when their diets are sharply restricted and then enrich

natka813 [3]

Can you make this question clearer to read?

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3 years ago

What is 119.6<br><img src="https://tex.z-dn.net/?f=%20%5Cdiv%20" id="TexFormula1" title=" \div " alt=" \div " align="absmiddle"

leva [86]

Answer: 39.866

That’s the answer even doe it was easy n u could of just done it

3 0

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!

4 0

3 years ago