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3 years ago

Which expression is equivalent to 60x^20 y^24/30x^10 y^12?

Mathematics

1 answer:

photoshop1234 [79]3 years ago

5 0

The answer to your equation is A. 2x^2y^2

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A triangle has a perimeter of 165 cm. The first side is 65 cm less than twice the second side. The third side is 10 cm less than

Neko [114]

Answer:A triangle has a perimeter of 165 cm. The first side is 65 cm less than twice the second side. The third side is 10 cm less than the second side. Write and solve an equation to find the length of each side of the triangle.

The sides can be found by taking the square root of the area. , where = side. . So the length of a side is 6.3 cm.

Step-by-step explanation:

3 0

3 years ago

Find the missing coefficient. - y2 − [-5y − y(-7y − 9)] − [-y (15y + 4)] = 0

maria [59]

There is only one solution in the given equation -y2 − [-5y − y(-7y − 9)] − [-y (15y + 4)] = 0. In solving this problem, apply first PEMDAS (parenthesis, exponents,multiplication, division, addition, subtraction). Then equation will transform into -y2+5y-7y2-9y+15y2+4y=0. Combine terms with same power and achieve 7y2=0. Divide both sides with 7 and perform square root of zero. Since the root is zero, we have one solution of the given equation which is y=0.

5 0

3 years ago

Read 2 more answers

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

3 years ago

What is the quotient of (3x^4 - 4x^2 + 8 - 1) ÷ (x - 2)? Thank God and Jesus Christ for another day and year of life❤✝️! Don't f

arsen [322]

Answer:

The answer is c

Step-by-step explanation:

cuzzzzzzz I toooooookkkkk aaaa test

6 0

3 years ago

a plane flying with the wind travels 630 mi in 1.5hrs. Flying against the wind , the plane travels 1120 mi in 4hrs. Find the rat

Flauer [41]

$\bf \begin{cases}
r=\textit{rate of the plane}\\
w=\textit{rate of the wind}\\
\end{cases}
\\\\

\begin{array}{ccccllll}
&distance&rate&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
\textit{with wind}&630&r+w&1\frac{1}{2}\\
\textit{against wind}&1120&r-w&4
\end{array}
$

$\bf thus \begin{cases}
630=(r+w)1\frac{1}{2}\to 630=(r+w)\frac{3}{2}
\\\\
1120=(r-w)4\\
--------------\\
630=(r+w)\frac{3}{2}\implies 630\cdot \frac{2}{3}=r+w\\
420=r+w\implies \boxed{420-w}=r\\
--------------\\
thus\\
--------------\\
1120=(r-w)4\implies 1120=4r-4w
\\\\
1120=4(\boxed{420-w})-4w
\end{cases}
$

solve for "w", to find the wind's speed rate,

so hmmm what's the plane's rate? well 420 - w = r :)

5 0

3 years ago