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devlian [24]
3 years ago
10

True or false. Y=3xsquared (like the little number two above the x) is direct variation.

Mathematics
1 answer:
galina1969 [7]3 years ago
3 0
Your answer will be true
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Someone please tell if this is right!
sertanlavr [38]

Answer:

Step-by-step explanation:

-4x , for the top right box, the rest are good

4 0
2 years ago
Circle the quadrilateral that does not belong with the other three. Explain your reasoning please help me this is due tomorrow
jeka94
Trapezoid because it doesn't have to parareles sides like the other
3 0
3 years ago
Find a normal vector n to the plane z−5(x−2)=2(8−y)
Hitman42 [59]

A normal vector is the set of coefficients of x, y, and z when the equation is written in standard (or general) form.

Subtracting the left side, we have

... 2(8 -y) -z +5(x -2) = 0

... 16 -2y -z +5x -10 = 0

... 5x -2y -z +6 = 0 . . . . . general form

A normal vector is (5, -2, -1).

7 0
3 years ago
Geometry please help!:(
Burka [1]
It would be triangle because you're cutting it
5 0
3 years ago
Calculus: Help ASAP
Serggg [28]
\bf \displaystyle \int\limits_{-1}^{0}~(4x^6+2x)^3(12x^5+1)\cdot  dx\\\\
-------------------------------\\\\
u=4x^6+2x\implies \cfrac{du}{dx}=24x^5+2\implies \cfrac{du}{2(12x^5+1)}=dx\\\\
-------------------------------\\\\
\displaystyle \int\limits_{-1}^{0}~u^3\underline{(12x^5+1)}\cdot\cfrac{du}{2\underline{(12x^5+1)}}\implies \cfrac{1}{2}\int\limits_{-1}^{0}~u^3\cdot du\\\\
-------------------------------\\\\

\bf \textit{now, we'll change the bounds, using u(x)}
\\\\\\
u(-1)=4(-1)^6+2(-1)\implies u(-1)=2
\\\\\\
u(0)=4(0)^6+2()\implies u(0)=0\\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{2}^{0}~u^3\cdot du\implies \left. \cfrac{1}{2}\cdot \cfrac{u^4}{4}  \right]_{2}^{0}\implies \left. \cfrac{u^4}{8}  \right]_{2}^{0}\implies [0]-[2]\implies -2
5 0
3 years ago
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