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Marina86 [1]
3 years ago
5

17

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
7 0

Answer:

The expressions that give the value of y are A - 3B and (1/3)A - B

The solution is (27/13, -60/13)

Step-by-step explanation:

We can see both equation A and equation B.

Equation A: 2x + (1/4)y = 3

Equation B: (2/3)x - y = 6

To find the value of y, we have to solve both equations A and equation B simultaneously. This is done by multiplying equation B by 3 and subtracting from equation A (A - 3B) to get:

(13/4)y = -15

y = -60/13

you can also get y by dividing equation A by 3 and subtracting equation B (1/3A - B)

Put y = -60/13 in equation A to get x:

2x + (1/4)(-60/13) = 3

2x = 3 + 15/13

2x = 54/13

x = 27/13

The solution is (27/13, -60/13)

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Solve by elimination. 1. 2x-3y=-9. 4x+9y=-13. 2. 2x-y=7 4x+5y=7. 3. 2x+4y=14 4x+y=14. 4. 4x-3y=16 5x+17y=20
JulijaS [17]
1.) x = -4, and y = 1/3

2.) x = 3, and y = -1

3.) x = 3, and y = 2

4.) x = 4, and y = 0

Hope this helps! - Alyssa
3 0
3 years ago
What is the answer for y= 6x+8
goldenfox [79]

Answer:

x=-1/4 thats the answer alternative

6 0
3 years ago
Find the volume of a pyramid with a square base, where the area of the base is
mestny [16]

Answer:

113.9 ft³

Step-by-step explanation:

==>Given:

Base area of square pyramid (B) = 17 ft²

Height of pyramid (h) = 20.1 ft

==>Required:

Volume of the square pyramid (V)

==>Solution:

Using the formula ⅓*B*h, let's find the volume of the pyramid as follows by plugging in our values:

V = ⅓*17*20.1

V = 341.7/3

V= 113.9 ft³

Volume of the square pyramid = 113.9 ft³

4 0
3 years ago
How do you factor 18r^5p^3-12r^3p^5+30rp^4
lozanna [386]
6rp^3 is the greatest common factor. Take that out and see what's left.

= 6rp^3*(3r^4 -2r^2p^2 +5p) . . . . . factors no further
3 0
4 years ago
A bird has a mass of 26 g and perches in the middle of a stretched telephone line. (a) Show that the tension in the line can be
Semmy [17]

Answer:

a) is demonstrated below, b) T=1.462N, c) T=14.6N

Step-by-step explanation:

a) Refer to the attached diagram.

Since the bird is standing in the middle of the line and each half is a straight line, Ta=Tb, so we will call the tension T=Ta=Tb and Tay=Tby

By trigonometry Tay=Ta·Sinθ

Since the system is in equilibrium W=Tay+Tby then:

W=2·Tay=2·Ta·Sinθ=2·T·Sinθ

Since W=mg, being m the mass of the bird and g, gravity:

mg=2TSin\theta

Isolating T, we demonstrate that

T=\frac{mg}{2Sin\theta}

b) Replacing θ=5º, m=0.026kg and g=9.8m/s² in the last equation, we can get the tension in Newtons:

T=\frac{0.026*9.8}{2Sin5}=1.462N

c) With θ=0.5º

T=\frac{0.026*9.8}{2Sin0.5}=14.6N

7 0
3 years ago
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