MATHEMATICS HIGH SCHOOL

Which of the following sets could be the sides of a right triangle?{2, 3,square root of 13 }
{2, 2, 4}
(1, 3,square root of 3 }

Answers

Answer 1
Answer:

Answer:

1st answer

Step-by-step explanation:

From pythagorus theorem.

Since hypotenuse

= \sqrt{ {2}^(2) + {3}^(2) }= \sqrt{ {13}}

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A circle is centered at the point (-3, 2) and passes through the point (1, 5). The radius of the circle is ______ units. The point (-7, ______) lies on this circle.How would I do this?

Answers

circle formula
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius

to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D=\sqrt{(x2-x1)^(2)+(y2-y1)^(2)}
points (-3,2) and (1,5)
D=\sqrt{(1-(-3))^(2)+(5-2)^(2)}
D=\sqrt{(4)^(2)+(3)^(2)}
D=√(16+9)
D=√(25)
D=5

center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1

the point (-7,5) and (7,-1) lie on this circle



radius=5 units
the points (-7,5) and (-7,1) lie on this circle






A pulley with a radius of 8 inches rotates three times every five seconds. Find the
angular velocity of the pulley in radians/sec (round to the nearest hundredth). Find the
linear velocity to the nearst ft/hr. 

Answers

If the pulley rotates at a rate of 3 revolutions per second, then the period T of movement is  (1)/(3)s

a) calculate the angular velocity:

\omega=(2 \pi)/(T)\n \n \omega=(2 \pi)/((1)/(3))=6 \pi \ rad/s

b) calculate the linear velocity:

v=(2 \pi R)/(T)=(2 \pi.8)/((1)/(3))=24 \pi \ in/s \approx 75,36 \ in/s

Remember: 1 in/s = 300 ft/h

So, 75,36 in/s = 22,608 ft/h

Find the mean, median, and interquartile for the data set below. Round your answer to the nearest tenth. 27,16,8,5,19,14,22,31,13,5,23,16,8

Answers

Answer:

Hence,

Mean=15.9

Median=16

Interquartile=14.5

Step-by-step explanation:

On arranging our data in the increasing order we get the data set as:

5   5   8   8   13   14   16    16    19    22   23   27   31

Now we find the mean as:

Mean=(5+5+8+8+13+14+16+16+19+22+23+27+31)/(13)\n\nMean=(207)/(13)\n\nMean=15.92

Now, we know that the median is the central tendency of the data and lie in the middle of the data.

Hence, by looking at the data we see that the middle value of the data is 16.

Hence, Median=16.

Now, the lower set of data is:

5   5   8   8  13   14

Now, the lower quartile is calculated as:

The middle value of the lower set of data is:

8     8

Hence, the lower quartile ]Q_1 is middle of 8,8.

i.e. (8+8)/(2)=8

similarly the upper set of data is:

16    19    22   23   27   31.

Similarly the upper quartile is calculated as:

Q_3=(22+23)/(2)=22.5

Hence, interquartile range (IQR) is calculated as:

Q_3-Q_1=22.5-8=14.5

Hence,

Mean=15.9

Median=16

Interquartile=14.5
The mean is 15.92, the median is 16, and the interquartile range is 14.5

Given ƒ(x) = 4 − x and g(x) = −4x, find (ƒ ○ g)(x)(ƒog) (x) = x
(ƒog) (x) = 4 - 4x
(ƒog) (x) = 4 + 4x
(ƒog) (x) = 0

Answers

f(x)=4-x
g(x)=-4x

(f o g)(x)=f(g(x))=4-(-4x)=4+4x

Answer: (fog)(x)=4+4x

In ΔOPQ, p = 180 inches, q = 120 inches and ∠O=171°. Find the length of o, to the nearest inch.

Answers

Given:

In ΔOPQ, p = 180 inches, q = 120 inches and ∠O=171°.

To find:

The length of o, to the nearest inch.

Solution:

According to cosine formula,

a^2=b^2+c^2-2bc\cos A

Using cosine formula in ΔOPQ, we get

o^2=p^2+q^2-2pq\cos O

On substituting the values, we get

o^2=(180)^2+(120)^2-2(180)(120)\cos (171^\circ)

o^2=32400+14400-43200(-0.9877)

o^2=46800+42668.64

o^2=89468.64

Taking square root on both sides.

o=√(89468.64)

o=299.113

o\approx 299

Therefore, the length of o is about 299 inches.

Answer:299

Step-by-step explanation:

Which one is it please hurry. Thank you

Answers

Answer: I think it’s A if I am incorrect I am very sorry

Step-by-step explanation:
I think it’s A but sorry if I’m wrong
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