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2 years ago

If you guys are in K12 and are in 7th grade help me!!!!!!!!!!!

Mathematics

1 answer:

Alex Ar [27]2 years ago

8 0

I wish I could help u with this one

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rick won 27 tickets playing a game. to get a prize, he needs to win at least 30 tickets. how many more tickets can he win to get

Svetach [21]

He needs 3 Tickets to Win a Prize.27×2 is... And that's ur Answer

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2 years ago

I need some help on this, please.

Free_Kalibri [48]

Hey there!

Area= length*width.

6.2 is the length and 3.7 is the width

3.7 *6.2= 22.94 yd^2

I hope this helps!

~kaikers

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2 years ago

Complete the square to rewrite y = x2 - 6x + 3 in vertex form. Then state

Vladimir [108]

Answer:

see explanation

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Given

y = x² - 6x + 3

To complete the square

add/subtract ( half the coefficient of the x- term)² to x² - 6x

y = x² + 2(- 3)x + 9 - 9 + 3

= (x - 3)² - 6 ← in vertex form

• If a > 0 then vertex is a minimum

• If a < 0 then vertex is a maximum

Here a = 1 > 0, thus

(3, - 6 ) is a minimum

3 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago

CAN YOU HELP ME ND WHO LOVE BILLIE ELISHIE

Alla [95]

I don’t like billies music but i like her as a person

8 0

2 years ago

Read 2 more answers