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Zinaida [17]
3 years ago
12

- Point K is on line segment JL. Given JK = 3 and KL = 8, determine the length JL.

Mathematics
1 answer:
AlekseyPX3 years ago
3 0

Answer:

JL = 11 units

Step-by-step explanation:

Add the units between JK and KL together to get the answer for JL

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Find the distance between points A and B.
VikaD [51]

Answer:

8 units

Step-by-step explanation:

you can just count the boxes and you get the answer.

7 0
3 years ago
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carol is excited to get a new kitten from the animal shelter. the kitten had an $299 adoption fee. carol had $50 off coupon for
Molodets [167]

Answer:

1 month

Step-by-step explanation:

249 is the amount she spends for buying the kitten, (299-50=249) which she only has to pay once.

20 is the amount she has to spend every month, the rate of change.

The problem can be modeled with the general equation:

y = 20x + 249

In this equation, x the the time in months and y is the amount of money spent.

Substitute 250 for y.

y = 20x + 249

250 = 20x + 249    <= isolate x to get the number of months

1 = 20x

x = 1/20

Carol will only spent money every month, not for 1/20 of a month.

It will only take Carol 1 month to spent $250.

Check answer:

Substitute x for 1

y = 20x + 249

y = 20(1) + 249

y = 269

269 is more than 250 already.

3 0
3 years ago
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Rudik [331]
He should change the coefficient 5
4 0
3 years ago
Factoring quadratics 4x^2-13x+3=0
ozzi

Answer:

X=3,1/4

Step-by-step explanation:

4x²-(12+1)x+3=0

or,4x²-12x-x+3=0

or,4x(x-3)-1(x-3)=0

or,(4x-1)(x-3)=0

Either,4x-1=0➡️ 4x=1

X=1/4

OR,x-3=0➡️ X=3

6 0
3 years ago
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Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 5 sin2 t, y = 5 cos2 t
rodikova [14]
\begin{cases}x(t)=5\sin2t\\y(t)=5\cos2t\end{cases}\implies\begin{cases}\frac{\mathrm dx}{\mathrm dt}=10\cos2t\\\frac{\mathrm dy}{\mathrm dt}=-10\sin2t\end{cases}

The distance traveled by the particle is given by the definite integral

\displaystyle\int_C\mathrm dS=\int_0^{3\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

where C is the path of the particle. The distance is then

\displaystyle\int_0^{3\pi}\sqrt{100\cos^22t+100\sin^22t}\,\mathrm dt=10\int_0^{3\pi}\mathrm dt=30\pi
6 0
3 years ago
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