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alexandr402 [8]
3 years ago
5

SOMEONE HELP ME PLEASE

Mathematics
1 answer:
IgorC [24]3 years ago
3 0

A standard die has numbers = 1,2,3,4,5 and 6

Total outcomes = 6

P = Number of outcomes/Total Outcomes

Even and less than 4 = 1,2,3,4 and 6

P = 5/6

Answered by GauthMath if you like please click thanks and comment thanks

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I need help writing the equation for the cost per packet and the cost per sheet
Goshia [24]
The answer to question A is 0.18 B 3.75
8 0
3 years ago
PLEASE HELPPP!!!
Tems11 [23]

Answer:

a) -16t^{2}+64t\geq40

b)[2-\frac{\sqrt{6}}{2} , 2+\frac{\sqrt{6}}{2}]

(see attached picture for the number line)

This interval means that from about 0.78s to about 3.22s the ball will not be visible because it will be higher than 40 feet high.

c) h_{max}=64ft

d) t=2s

Step-by-step explanation:

a)

Basically the ball will disappear when it is higher than 40 ft, so we can build the inequality like this:

-16t^{2}+64t\geq40

this is because the function represents it's height, so when the function is greater than or equal to 40, then the ball will disappear.

b) to solve the inequality, first we need to turn the ≥ symbol to an = symbol and solve for t, so we get:

-16t^{2}+64t =40

we can move the 40 to the other side of the equation so we get:

-16t^{2}+64t-40=0

we can factor the equation so it's easier to solve, so we get:

-8(2t^{2}-8t+5) =0

which simplifies to:

2t^{2}-8t+5=0

and we can use the quadratic formula to solve this equation, so we get:

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

so we get:

t=\frac{-(-8)\pm\sqrt{(-8)^{2}-4(2)(5)}}{2(2)}

which yields:

2\pm \frac{\sqrt{6}}{2}

and next we need to test the posible intervals to see which one makes the inequality true, the possible intervals are:

(-\infty, 2-\frac{\sqrt{6}}{2}]  for a test value of 0

[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}] for a test value of 1

[2+\frac{\sqrt{6}}{2},\infty) for a test value of 4

so next we test each of the values in the original inequality and see if the inequality is true:

(-\infty, 2-\frac{\sqrt{6}}{2}] for a test value of 0

-16(0)^{2}+64(0)\geq40

0\geq40 false

so this is not an answer.

[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}] for a test value of 1

-16(1)^{2}+64(1)\geq40

48\geq40

true, so this is an answer.

[2+\frac{\sqrt{6}}{2}, \infty) for a test value of 4

-16(4)^{2}+64(4)\geq40

-16(16)+64(4)\geq40

-256+256\geq40

0\geq40

false, so this is no answer.

So the answer is the interval:

[2-\frac{\sqrt{6}}{2},2+\frac{\sqrt{6}}{2}]

This interval means that from about 0.78s to about 3.22s the ball will not be visible because it will be higher than 40 feet high.

c) In order to find how high Penelope kick the football, we must find the vertex of the parabola which is gotten when graping the given function (see attached picture)

so we can use the following formula to find the t-value for this maximum height:

t=-\frac{b}{2a}

so we substitute the corresponding values:

t=-\frac{64}{2(-16)}

which yields:

t=2s

next, we can substitute this time into the original function to find the maximum height, so we get:

h=-16(2)^{2}+64(2)

which yields:

h=64 ft

d) We found the answer for d on the previous part where we got that t=2s

6 0
3 years ago
Delia has three pieces of ribbon. Her red ribbon is 2/4 foot long. Her green ribbon is 2/3 foot long. Her yellow ribbon is 2/6 f
Scorpion4ik [409]
The green ribbon is the correct ansser
7 0
3 years ago
Read 2 more answers
HELP!!
Leokris [45]

Answer:

1/2

Step-by-step explanation:

Given:

sinx=-1/2

cosy=√3/2

finding x

x=sin^-1(-1/2)

x=-π/6

x=-30°

finding y

cosy=√3/2

y=cos^-1(√3/2)

y=π/6

y=30°

Now finding cos(x-y)

cos(-π/6-π/6)

=cos(-π/3)

=1/2!

6 0
3 years ago
Can you please help with both? :)​
QveST [7]

Answer: 1. 12 2. 120

Step-by-step explanation:

1. so basically 9/6=18/x

since they are similir the'd have the same ratio

sooo you just solve for x (multiply 6 by 18, and then divide by 9) and you get 12

2. same thing except you put 6/10=12/x, and solve it (multiply 10 by 12 and divide by 6) and u get 20

dats how i would do it but there are other ways too im sure :)

hope this helps

8 0
2 years ago
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