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viva [34]
3 years ago
10

What is the answer 45w + 123.95 = 753.95

Mathematics
2 answers:
andrey2020 [161]3 years ago
8 0

subtract 123.95 from both sides

45w = -630

divide both sides by 45

w = -14

expeople1 [14]3 years ago
3 0

Answer:

the answer is 14 your welcome :)

Step-by-step explanation:

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What is the inverse of the function f(x) = 2x – 10?
Evgesh-ka [11]
First make x the subject of f(x) = 2x - 10:-

2x = f(x) + 10

x =  1/2 f(x) + 5

mow replace x by hx) and f(x) by x:-

h(x )= 1/2 x + 5
3 0
3 years ago
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Explain why it is necessary to rename 4 1/2 if you subtract 3/4 from it
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3 years ago
A chord is 10cm from the centre of a circle of radius 15cm. Find the length of the chord.
almond37 [142]

Answer: 21.44 cm

Step-by-step explanation:

Given

Radius r=15 cm

distance of chord from the center is 10 cm

From the figure, the length of the chord is AC

Applying Pythagoras

\Rightarrow AO^2=OB^2+AB^2\\\Rightarrow AB^2=15^2-10^2\\\Rightarrow AB^2=225-100=115\\\Rightarrow AB=\sqrt{115}=10.72\ cm

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5 0
3 years ago
1. A right triangle is shown below. Use the Pythagorean Theorem to determine the lengths of each side. Phythagorean Theorem: a2
erik [133]
This may not be very pretty.
(2x + 5) * (2x + 5) expanded is (use foil)
F = 4x^2
O = 2x * 5 = 10x
I = 2x + 5 = 10x
L = 25
Total = 4x^2 + 20x + 25

(x + 3)(x + 3) = x^2 + 6x + 9 by the same method

 (2x + 7)(2x + 7) = 4x^2 + 28x + 49 Same method.

4x^2 + 20x + 25 + x^2 + 6x + 9 = 4x^2 + 28x + 49
5x^2 + 26x + 34 = 4x^2 + 28x + 49 Collect everything on the left.
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x + 3 will have no meaning.
x - 5 = 0
x = 5

2x + 5 = 2*5 + 5 = 15
x + 3 = 5 + 3 = 8
2x+ 7 = 2*5 +7 = 17

Check
15^2 + 8^2 = 17^2
225 + 64 = ? 289
289 = 289

a = 15 <<<<< answer
b = 8 <<<<< answer
c = 17 <<<<< answer
 


5 0
3 years ago
PLEASE HURRRYY!!! I need help
VashaNatasha [74]

Answer:

C) 8 ft

Step-by-step explanation:

you use the pythagorean theorem to find the height.

a² + b² = c²

a = radius

b = height

c = slant height

15² + h² = 17²

h = 8

3 0
3 years ago
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