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Marina CMI [18]
1 year ago
12

Solve for X and Y: Y=2+7x and -2x+6y=-28

Mathematics
1 answer:
Furkat [3]1 year ago
6 0

Answer:

x=-1 and y=-5

Step-by-step explanation:

ሰላም!

Given expressions let them given as two equations

  • y=2+7x....... equation (1)
  • -2x+6y=-28..... equation (2)

Thus, substitute y=2+7x into equation (2)

- 2x + 6(2 + 7x) =  - 28 \\

simplify it

- 2x + 12 + 42x =  - 28 \\ 40x + 12 =  - 28

subtract 12 from both sides

40x + 12 - 12 =  - 28 - 12 \\ 40x =  - 40

divide both sides by 40

\frac{40x}{40}  =  \frac{ - 40}{40} \\ x =  - 1

Therefore, as we get the value of x=-1 substitute it to equation (1) and solve for variable y

y = 2 + 7x \\ y = 2 + 7( -1 ) \\ y = 2 - 7 \\ y =  - 5

To be more sure make a cross check on both of the two expressions

Thus,

1) y=2+7x .... substitute y=-5 and x=-1 and check if it is equal

-5=2+7(-1)

-5=2-5

-5=-5

2)-2x+6y=-28... substitute and check

-2(-1)+6(-5)=-28

2-30=-28... (-a)(-b)=ab

-28=-28

Hope it helps!

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(a) Find the unit tangent and unit normal vectors T(t) and N(t).
andrey2020 [161]

Answer:

a. i. (i + tj + 2tk)/√(1 + 5t²)

ii.  (-5ti + j + 2k)/√[25t² + 5]

b. √5/[√(1 + 5t²)]³

Step-by-step explanation:

a. The unit tangent

The unit tangent T(t) = r'(t)/|r'(t)| where |r'(t)| = magnitude of r'(t)

r(t) = (t, t²/2, t²)

r'(t) = dr(t)/dt = d(t, t²/2, t²)/dt = (1, t, 2t)

|r'(t)| = √[1² + t² + (2t)²] = √[1² + t² + 4t²] = √(1 + 5t²)

So, T(t) = r'(t)/|r'(t)| = (1, t, 2t)/√(1 + 5t²)  = (i + tj + 2tk)/√(1 + 5t²)

ii. The unit normal

The unit normal N(t) = T'(t)/|T'(t)|

T'(t) = dT(t)/dt = d[ (i + tj + 2tk)/√(1 + 5t²)]/dt

= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + [-10tk/√(1 + 5t²)⁻³]

= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + j/√(1 + 5t²)+ [-10t²k/√(1 + 5t²)⁻³] + 2k/√(1 + 5t²)

= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) - 10t²k/[√(1 + 5t²)]⁻³ + 2k/√(1 + 5t²)

= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ - 10t²k/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) + 2k/√(1 + 5t²)

= -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + (j + 2k)/√(1 + 5t²)

We multiply by the L.C.M [√(1 + 5t²)]³  to simplify it further

= [√(1 + 5t²)]³ × -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + [√(1 + 5t²)]³ × (j + 2k)/√(1 + 5t²)

= -(i + tj + 2tk)5t + (j + 2k)(1 + 5t²)

= -5ti - 5²tj - 10t²k + j + 5t²j + 2k + 10t²k

= -5ti + j + 2k

So, the magnitude of T'(t) = |T'(t)| = √[(-5t)² + 1² + 2²] = √[25t² + 1 + 4] = √[25t² + 5]

So, the normal vector N(t) = T'(t)/|T'(t)| = (-5ti + j + 2k)/√[25t² + 5]

(b) Use Formula 9 to find the curvature.

The curvature κ = |r'(t) × r"(t)|/|r'(t)|³

since r'(t) = (1, t, 2t), r"(t) = dr'/dt = d(1, t, 2t)/dt  = (0, 1, 2)

r'(t) = i + tj + 2tk and r"(t) = j + 2k

r'(t) × r"(t) =  (i + tj + 2tk) × (j + 2k)

= i × j + i × 2k + tj × j + tj × 2k + 2tk × j + 2tk × k

= k - 2j + 0 + 2ti - 2ti + 0

= -2j + k

So magnitude r'(t) × r"(t) = |r'(t) × r"(t)| = √[(-2)² + 1²] = √(4 + 1) = √5

magnitude of r'(t) = |r'(t)| = √(1 + 5t²)

|r'(t)|³ = [√(1 + 5t²)]³

κ = |r'(t) × r"(t)|/|r'(t)|³ = √5/[√(1 + 5t²)]³

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2 years ago
Which shows the correct substitution of the values a, b, and c from the equation –2 = –x + x2 – 4 into the quadratic formula?
Serhud [2]
X=[-(-1)+or-sqrt ((-1)^2-4x1x-2)]/(2x1)

7 0
3 years ago
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At the end of the day of teaching the skill of cutting and sewing to make capes, Ms. Ironperson and Mr. Thoro decided to go to t
olga_2 [115]

Answer:

The cost of a chicken shawarma wrap = $10.99

The cost of an order of spiced potatoes=$4.99

Equations needed

3x+2y =42.95

5x + 4y=74.91

Step-by-step explanation:

x = the cost of a chicken shawarma wrap

y = the cost of an order of spiced potatoes

Ms. Ironperson

3x+2y =42.95

Mr. Thoro

5x + 4y=74.91

3x+2y =42.95 (1)

5x + 4y=74.91 (2)

Multiply (1) by 2

6x + 4y=85.9 (3)

5x + 4y=74.91. (2)

Subtract (2) from (3)

6x - 5x= 85.9 - 74.91

x=$10.99

Substitute the value of x into (1)

3x+2y =42.95 (1)

3(10.99) + 2y =42.95

32.97 + 2y = 42.95

2y = 42.95 - 32.97

2y=9.98

Divide both sides by 2

y=9.98/2

=4.99

y=$4.99

Therefore,

The cost of a chicken shawarma wrap = $10.99

The cost of an order of spiced potatoes=$4.99

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3 years ago
Which set of ordered pairs represents a function? {(0, 1), (1, 3), (1, 5), (2, 8)} {(0, 0), (1, 2), (2, 6), (2, 8)} {(0, 0), (0,
11111nata11111 [884]

Answer: {(0, 2), (1, 4), (2, 6), (3, 6)}

Step-by-step explanation:

For a relation to be considered a function, each x-value needs to have one corresponding y-value--it cannot have more than 1.

Since all the other sets of ordered pairs feature points with two x-values with different y-values, the set above is the only function of the provided options.

4 0
3 years ago
60 percent of what is 181
dangina [55]
181(0.60)=108.6 or 543/5
4 0
3 years ago
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