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Nesterboy [21]
3 years ago
14

Square root:27 + SR:48+SQ:12 = ???

Mathematics
2 answers:
Anika [276]3 years ago
5 0
Sqrt(27)+sqrt(48)+sqrt(12)
Sqrt(27)=sqrt(3*3*3)=3*sqrt(3)
Sqrt(48)=sqrt(4*4*3)=4*sqrt(3)
Sqrt(12)=sqrt(2*2*3)=2*sqrt(3)
Add the three to get 9*sqrt(3)
Final answer:
9*sqrt(3)
Hope I helped :)
Nana76 [90]3 years ago
4 0
12 + 2 = 14..........................................................

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Lena [83]

All we need to do is plug the numbers into the equation.


A:

2(3)--1=0\\6--1=0\\7\neq0

Not A.

B:

2(2)--4=0\\4--4=0\\8\neq0

Not B.

C:

2(1)-2=0\\0=0

C is the correct answer

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Find the median of each set of data. 12, 8, 6, 4, 10, 1 7 6, 3, 5, 11, 2, 9, 5, 0 5 30, 16, 49, 25
professor190 [17]

Answer:

first 7 then 5 and 27.5

Step-by-step explanation:

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3 years ago
2/3 (t+5) = 8<br><br>what does t equal?​
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6 0
3 years ago
marsha waide borrows 5,000 and agrees to pay it back in 2 years . if the simple interst rate is 13% ,find the total amount she p
Dmitry_Shevchenko [17]

Answer:

You want to calculate the interest on $5000 at 13% interest per year after 2 year(s).

The formula we'll use for this is the simple interest formula.

Where:

P is the principal amount, $5000.00.

r is the interest rate, 13% per year, or in decimal form, 13/100=0.13.

t is the time involved, 2....year(s) time periods.

So, t is 2....year time periods.

To find the simple interest, we multiply 5000 × 0.13 × 2 to get that:

The interest is: $1300.00

Usually now, the interest is added onto the principal to figure some new amount after 2 year(s),

or 5000.00 + 1300.00 = 6300.00. For example:

If you borrowed the $5000.00, you would now owe $6300.00

8 0
3 years ago
An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(b)Flight Direction of Y

Using Law of Sines

\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)

The direction of flight Y to the nearest degree is 39 degrees.

7 0
3 years ago
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