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3 years ago

Are points K, X, Q and L collinear or coplanar?

1 answer:

3 0

Could use a diagram, but just remember collinear points all lie on a straight line, in the three dimensional world, when a set of points all lie on the same plane, they are called coplanar.

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use any strategy to solve the following problems 1) for haloween ned received 48 pieces of candy .if he put them into piles with

Veronika [31]

Answer:

He could make 16 piles.

Step-by-step explanation:

Divide 48 by 3: 48 ÷ 3 = 16

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3 years ago

Why was formal education more important fo urban americans than rural american?

jasenka [17]

Most higher-paying jobs in the cities required a formal education. Hope it helps you !

5 0

4 years ago

Read 2 more answers

Factor -24a3B3 - 84a4bc

Eduardwww [97]

(−12a^3b)(7ac+2b^2)

Step-by-step explanation:

(−24a^3(b^3)−84a^4bc

−84a^4bc−24a^3b^3=

(−12a^3b)(7ac+2b^2)

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3 years ago

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Temka [501]

Answer:

What?

Step-by-step explanation:

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3 years ago

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Solve the initial value problem

tigry1 [53]

$9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}$

Look for an integrating factor $\mu(t)$ :

$\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}$

Multiply both sides by $\mu$ :

$(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}$

Condense the left side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}$

Integrate both sides:

$(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt$

For the integral on the right, substitute

$u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du$

$\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du$

$\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C$

$\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)$

$\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C$

$\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}$

Given that $y(0)=12$ , we get

$12=9+C\implies C=3$

$\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}$

6 0

3 years ago