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Otrada [13]
3 years ago
12

If Q = 53 cm and P = 28 cm , what is the length of R?

Mathematics
2 answers:
Alchen [17]3 years ago
4 0
To get the answer we can use Pythagorean theroem.
We know that P^2+R^2=Q^2     /-P^2
R^2=Q^2-P^2
Now we can substitute our data.
R^2=53^2+28^2 \\ R^2=3592 \\ R=60 - its the answer
Murljashka [212]3 years ago
4 0

Answer:

The answer is C. 45cm

Step-by-step explanation:

Use the Pythagorean theorem to find the length of r.

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Solve for R. 3 + 3r > 9
Nostrana [21]

Answer:

r > 2

Step-by-step explanation:

9-3 = 6

6/3 = 2

Have a good day :)

7 0
3 years ago
What's the vertex for 6x^2-48x-54=0?
kakasveta [241]
It would be 4,-25 becuase of in the first place i in and the other imp in cams was the
4 0
2 years ago
Someone please help me find y.
WITCHER [35]
Here's y

y=<span>‌<span><span><span>−<span>8x</span></span>+29
       </span><span><span>−x</span>+<span>3

Hope this helps:)
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7 0
3 years ago
A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)= at² + bt + c, where t is the
frutty [35]

Answer:

<h2>a) a = -3, b = 18, c = 48; </h2><h2>s(t) = -3t²+18t+48</h2><h2>b) 48m</h2><h2>c) 8secs</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

'A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)=at²+bt+c, where t is the time in seconds after the rock was released. After 1 second the rock was 63 m above sea level, after 2 seconds 72 m, and after 7 seconds 27 m. a. Find a, b and c and hence an expression for s(t). b. Find the height of the cliff. c. Find the time taken for the rock to reach sea level.'

Given the equation of the distance modelled as s(t)=at²+bt+c

If after 1 second the rock was 63 m, then 63 = a+b+c

If after 2 seconds, the distance was 72 m then 72 = 4a+2b+c

Also if after 7 seconds, the distance is 27 m, then 27 = 49a+7b+c

i) Solving the 3 equations simultaneously to get a, b and c we have;

a+b+c = 63 ... 1

4a+2b+c = 72 ...2

49a+7b+c = 27 ...3

Subtracting 2 from 1 and 3 from 2 we will generate 2 new equations as shown;

eqn 2- eqn1: 3a + b = 9...4

eqn 3- eqn 2: 45a + 5b = -45

eqn 3- eqn 2: 9a+b = -9 ... 5

solving 4 and 5 simultaneously

3a + b = 9 ...4

9a+b = -9 ... 5  

Taking the difference of 4 and 5 we have

6a - -9-9

6a = -18

a = -3

substituting a = -3 into equation 4 to get b we have;

3(-3)+b = 9

-9 + b = 9

b = 9+9

b = 18

substituting a = -3 and b = 18 into equation 1 to get c we have;

-3+18+c = 63

15+c = 63

c = 48

a = -3, b= 18 and c = 48

The distance function will be s(t) = -3t²+18t+48

ii) If the height of the cliff is modelled by the equation  s(t)=at²+bt+c

The height of the cliff is at when t = 0

s(0) = -3(0)²+18(0)+48

s(0) = 48m

The height of the cliff is 48m

iii) At the sea level, the height of the rock will be 0m, substituting this into the modeled equation for the height to get the time we have;

s(t)=at²+bt+c

0 = -3t²+18t+48

3t²-18t-48 = 0

t² - 6t - 16 =0

t² - 8t+2t - 16 = 0

t(t-8)+2(t-8) = 0

(t+2)(t-8) = 0

t = -2 or 8

Taking the positive value of the time, t = 8secs

Time taken for the rock to reach sea level is 8secs

7 0
3 years ago
A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student
Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that p = 0.25

Class of 18 students

This means that n = 18

What would be the probability that at least 6 students drive themselves to school?

This is

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096

P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17

P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213

P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282

Closest option is B, just a small rounding difference.

4 0
3 years ago
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