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2 years ago

What two fractions equal -3/5

Mathematics

1 answer:

Kaylis [27]2 years ago

7 0

Answer:

-6/10 and -12/10 (I think)

Step-by-step explanation:

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R= 4/3(p-q) solve for p

Naya [18.7K]

Answer = p=12-4 -12=17=18+19

p-q

p=12

3 0

2 years ago

Solve the equation -18 + 24 = -2(x+6)

anzhelika [568]

Answer:

x = -9

Step-by-step explanation:

-18+24 = -2(x+6) Distribute and simplify

6 = -2x-12 Add 12 to both sides

18 = -2x Divide by -2 on both sides

-9 = x Here's your answer

Hope this helps! :D

3 0

1 year ago

Read 2 more answers

What is the slope of these two equations?

Juli2301 [7.4K]

Answer:

8. 0

9. undefined

Step-by-step explanation:

8. 0

> because the y variable is the same for all x-values, this is a horizontal line. Horizontal lines have a slope of 0.

> Thinking of slope as rise over run: we will always rise 0, and run __ from any two points--0 divided by any number is always 0

9. undefined

> because the x variable is always the same, no matter what y variable we graph, we will have the same outcome. So, this would look like a straight line, which have an undefined slope.

> If you think of a slope as rise / run; if you go from any two points, there will be a 0 in the denominator--which is undefined

hope this helps!! have a lovely day :)

3 0

1 year ago

Please can someone help me with this question

sasho [114]

Answer:

r = - $\frac{6a}{1-5m^2}$

Step-by-step explanation:

Rewrite the equation as $\sqrt{x} 6a+r/5r$ = m

Remove the radical on the left side of the equation by squaring both sides of the equation.

( $\sqrt{\frac{6a+r}{5r} )^{2}$ = m^2

Then, you simplify each of the equation.

Rewrite: ( $\sqrt{\frac{6a+r}{5r} )^{2}$ as $\frac{6a+r}{5r} = m^{2}$

Remove any parentheses if needed.

Solve for r.

Multiply each term by r and simplify."

Multiply both sides of the equation by 5.

6a+r= m^2r⋅(5)

Remove parentheses.

Move 5 to the left of (m ^2) r

6a+r=5m^2)r

Subtract 5m^2)r from both sides of the equation.

6a+r-5m^2)r=0

Subtract 6a from both sides of the equation.

r-5m^2)r=-6a

Factor r out of r-5m^2)r

r(1-5m^2)=-6a

Divide each term by 1-5m^2 and simplify.

r = - $\frac{6a}{1-5m^2}$

There you go, hope this helps!

4 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago