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shusha [124]
3 years ago
14

Find the product (2x+5)^3 . Write your answer in standard form

Mathematics
1 answer:
seropon [69]3 years ago
5 0

Formula for this case is

(a+b)∧3=a∧3+3a∧2b+3ab∧2+b∧3

(2x+5)∧3=(2x)∧3+3*(2x)∧2*5+3*2x*5∧2+5∧3=8 x∧3+3*4x∧2*5+3*2x*25+125=

=8x∧3+60x∧2+150x+125

Good luck!!!

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Write a function for the situation. Is the graph continuous or discrete? A movie store sells DVDs for $11 each. What is the cost
Goshia [24]
B: because 11 times N will give you how much your cost would be . Example: 11*2= $22 if you bought 2 cds it would be $22. and its continuous cause youre going up at a constant rate.
7 0
3 years ago
The height of a triangle is twice the length of its base. The
lions [1.4K]

Answer:

height = 14.2 m , base = 7.1 m

Step-by-step explanation:

let the base be b then height h = 2b

The area (A) of a triangle is calculated as

A = \frac{1}{2} bh

Here A = 50 , then

\frac{1}{2} × b × 2b = 50

b² = 50 ( take square root of both sides )

b = \sqrt{50} ≈ 7.1 ( to the nearest tenth )

Then base = 7.1 m and height = 2 × 7.1 = 14.2 m

6 0
2 years ago
Ten times a number increased by eighty four equals 194. what is the number
steposvetlana [31]

Answer: x = 11

Step-by-step explanation:

10x+84=194

10x = "10 times a number"

+84 = "increased by eighty four"

=194 = "equals 194"

10x+84=194

Subtract 84

10x = 110

Divide by 10

x = 11

<em>Hope it helps <3</em>

4 0
3 years ago
Read 2 more answers
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
a parallelogram has one angle that measures 55 degrees. what are the measures of the other angles in the parrelogram.
Ugo [173]

a parallelogram has four sides so all its angles add up to make 360

the angle across from the 55 would also be a 55 equaling a 110 degrees angle leaving you with 250 so your other two angles both equal 125

3 0
3 years ago
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